fizikzauglom

GRADE 10 PRACTICE

Physics Problems: Dynamics, Grade 10

Newton's laws and fundamental forces.

32 problems across 4 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

32 problems
Position in this set: 1 from 32

Problem #35

Dynamics: Newton's laws

What force must be applied to a lead block of volume 40\,\text{см}^{3}so that it moves with acceleration 5\,\text{м/с}^{2}? How much will a dynamometer spring stretch? 120\,\text{Н/м}, if you hang this block from it? Lead density 11300 kg/m³; in the first experiment the applied force is the resultant one.

Answer options

  1. \left(22{,}6\,\text{Н};\ 0{,}377\,\text{м}\right)
  2. \left(2{,}26\,\text{Н};\ 0{,}0377\,\text{м}\right)
  3. \left(1{,}13\,\text{Н};\ 0{,}01885\,\text{м}\right)
  4. \left(4{,}52\,\text{Н};\ 0{,}0754\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
V
40\,\text{см}^{3}40\cdot 10^{-6}\,\text{м}^{3}
\rho
11{,}3\,\tfrac{\text{г}}{\text{см}^{3}}11\,300\,\tfrac{\text{кг}}{\text{м}^{3}}
a
5\,\tfrac{\text{м}}{\text{с}^{2}}
k
120\,\tfrac{\text{Н}}{\text{м}}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

F\Delta \ell

  1. m = \rho Vm = 11{,}3\,\tfrac{\text{г}}{\text{см}^{3}}\cdot 40\,\text{см}^{3}452\,\text{г} = 0{,}452\,\text{кг}
  2. F = maF = 0{,}452\,\text{кг}\cdot 5\,\tfrac{\text{м}}{\text{с}^{2}}F = 2{,}26\,\text{Н}
  3. F_{\text{т}} = mgF_{\text{т}} = 0{,}452\cdot 10F_{\text{т}} = 4{,}52\,\text{Н}

    The same block hangs on the spring, so it is stretched by gravity.

  4. F_{\text{у}} = k\Delta\ell\ \Rightarrow\ \Delta\ell = \dfrac{F_{\text{т}}}{k}\Delta\ell = \dfrac{4{,}52\,\text{Н}}{120\,\tfrac{\text{Н}}{\text{м}}}0{,}0377\,\text{м} \approx 3{,}8\,\text{см}

AnswerF = 2{,}26\,\text{Н}\Delta\ell = 0{,}0377\,\text{м}

Position in this set: 2 from 32

Problem #36

Dynamics: Newton's laws

Force imparts mass to a body m_1 acceleration a_1, and a body of mass m_2 - acceleration a_2. What acceleration will this force impart to a body of mass m_1+m_2?

Answer options

  1. a=\dfrac{a_1+a_2}{2},\ [a]=\text{м/с}^{2}
  2. a=\dfrac{a_1-a_2}{a_1+a_2},\ [a]=\text{м/с}^{2}
  3. a=\dfrac{a_1a_2}{a_1+a_2},\ [a]=\text{м/с}^{2}
  4. a=a_1+a_2,\ [a]=\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
m_1
m_1,\ a_1
m_2
m_2,\ a_2
Find

a

  1. m_1 = \dfrac{F}{a_1},\quad m_2 = \dfrac{F}{a_2}

    The force is the same, so we express the masses through it.

  2. m = m_1 + m_2 = \dfrac{F}{a_1} + \dfrac{F}{a_2} = \dfrac{F\left(a_1 + a_2\right)}{a_1 a_2}
  3. a = \dfrac{F}{m} = \dfrac{F\,a_1 a_2}{F\left(a_1 + a_2\right)}a = \dfrac{a_1 a_2}{a_1 + a_2}

Answera = \dfrac{a_1 a_2}{a_1 + a_2}

Position in this set: 3 from 32

Problem #37

Dynamics: forces and balance

On a sled with a load of mass 50\,\text{кг} horizontal force acts 40\,\text{Н}. With what acceleration does the sled move if the coefficient of friction is equal to 0{,}02?

Answer options

  1. 0{,}3\,\text{м/с}^{2}
  2. 1{,}2\,\text{м/с}^{2}
  3. 6\,\text{м/с}^{2}
  4. 0{,}6\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
m
50\,\text{кг}
F
40\,\text{Н}
\mu
0{,}02
Find

a

  1. F_{\text{тр}} = \mu m gF_{\text{тр}} = 0{,}02\cdot 50\cdot 10F_{\text{тр}} = 10\,\text{Н}
  2. F - F_{\text{тр}} = ma\ \Rightarrow\ a = \dfrac{F - F_{\text{тр}}}{m}a = \dfrac{40 - 10}{50}a = 0{,}6\,\tfrac{\text{м}}{\text{с}^{2}}

Answera = 0{,}6\,\tfrac{\text{м}}{\text{с}^{2}}

Position in this set: 4 from 32

Problem #38

Dynamics: Newton's laws

Force 4\,\text{Н} imparts acceleration to the body 1{,}25\,\text{м/с}^{2}. What force will impart acceleration to this body? 3\,\text{м/с}^{2}?

Answer options

  1. 9{,}6\,\text{Н}
  2. 4{,}8\,\text{Н}
  3. 19{,}2\,\text{Н}
  4. 96\,\text{Н}
Show the walkthroughHide the walkthrough
Given
F_1
4\,\text{Н}
a_1
1{,}25\,\tfrac{\text{м}}{\text{с}^{2}}
a_2
3\,\tfrac{\text{м}}{\text{с}^{2}}
Find

F_2

  1. F_1 = ma_1\ \Rightarrow\ m = \dfrac{F_1}{a_1}m = \dfrac{4\,\text{Н}}{1{,}25\,\tfrac{\text{м}}{\text{с}^{2}}}m = 3{,}2\,\text{кг}
  2. F_2 = ma_2F_2 = 3{,}2\,\text{кг}\cdot 3\,\tfrac{\text{м}}{\text{с}^{2}}F_2 = 9{,}6\,\text{Н}

    In short: F_1/F_2 = a_1/a_2, whence F_2 = F_1 a_2 / a_1.

AnswerF_2 = 9{,}6\,\text{Н}

Position in this set: 5 from 32

Problem #39

Dynamics: Newton's laws

With what acceleration does a rocket with a mass of 200\,\text{кг}, if the engine thrust is equal to 3\,\text{кН}?

Answer options

  1. 50\,\text{м/с}^{2}
  2. 5\,\text{м/с}^{2}
  3. 2{,}5\,\text{м/с}^{2}
  4. 10\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
m
200\,\text{кг}
F_{\text{т}}
3\,\text{кН}3000\,\text{Н}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

a

  1. F_{\text{т}} - mg = ma\ \Rightarrow\ a = \dfrac{F_{\text{т}} - mg}{m}

    Upward is the force of gravity, downward is the force of gravity.

  2. a = \dfrac{F_{\text{т}} - mg}{m}a = \dfrac{3000 - 200\cdot 10}{200}a = 5\,\tfrac{\text{м}}{\text{с}^{2}}

Answera = 5\,\tfrac{\text{м}}{\text{с}^{2}}

Position in this set: 6 from 32

Problem #40

Dynamics: forces and balance

With what force do the Earth and the Sun attract? Accept M_{\text{З}}=6\cdot10^{24}\,\text{кг}, M_{\text{С}}=2\cdot10^{30}\,\text{кг}, r=1{,}5\cdot10^{11}\,\text{м}. What diameter would a steel cable hold the Earth at its breaking point? 500\cdot10^{6}\,\text{Па}?

Answer options

  1. \left(7{,}12 \cdot 10^{22}\,\text{Н};\ 1{,}9 \cdot 10^{7}\,\text{м}\right)
  2. \left(3{,}56 \cdot 10^{23}\,\text{Н};\ 9{,}5 \cdot 10^{7}\,\text{м}\right)
  3. \left(3{,}56 \cdot 10^{22}\,\text{Н};\ 9{,}5 \cdot 10^{6}\,\text{м}\right)
  4. \left(1{,}78 \cdot 10^{22}\,\text{Н};\ 4{,}75 \cdot 10^{6}\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
M_{\text{З}}
6\cdot 10^{24}\,\text{кг}
M_{\text{С}}
2\cdot 10^{30}\,\text{кг}
r
150\cdot 10^{6}\,\text{км}150\cdot 10^{9}\,\text{м}
G
6{,}67\cdot 10^{-11}\,\tfrac{\text{Н}\cdot\text{м}^{2}}{\text{кг}^{2}}
\sigma_{\text{пр}}
500\cdot 10^{6}\,\text{Па}
Find

Fd

  1. F = G\dfrac{M_{\text{З}}M_{\text{С}}}{r^{2}}F = \dfrac{6{,}67\cdot 10^{-11}\cdot 6\cdot 10^{24}\cdot 2\cdot 10^{30}}{\left(150\cdot 10^{9}\right)^{2}}F = 3{,}56\cdot 10^{22}\,\text{Н}
  2. \sigma_{\text{пр}} = \dfrac{F}{S}\ \Rightarrow\ S = \dfrac{F}{\sigma_{\text{пр}}}S = \dfrac{3{,}56\cdot 10^{22}}{500\cdot 10^{6}}S = 7{,}1\cdot 10^{13}\,\text{м}^{2}
  3. S = \dfrac{\pi d^{2}}{4}\ \Rightarrow\ d = \sqrt{\dfrac{4S}{\pi}}d = \sqrt{\dfrac{4\cdot 7{,}1\cdot 10^{13}}{3{,}14}}d = 9{,}5\cdot 10^{6}\,\text{м} = 9500\,\text{км}

AnswerF = 3{,}56\cdot 10^{22}\,\text{Н}d = 9{,}5\cdot 10^{6}\,\text{м}

Position in this set: 7 from 32

Problem #41

Dynamics: forces and balance

Per body weight 6\,\text{кг} A horizontal force acts on a horizontal table. Take the coefficients of static and sliding friction to be equal 0{,}2. Find the force for uniform sliding and acceleration 1{,}5\,\text{м/с}^{2}. Then find the acceleration of the initially stationary body at forces 5\,\text{Н} and 15\,\text{Н}.

Answer options

  1. \left(6\,\text{Н};\ 10{,}5\,\text{Н};\ 0\,\text{м/с}^{2};\ 0{,}25\,\text{м/с}^{2}\right)
  2. \left(24\,\text{Н};\ 42\,\text{Н};\ 0\,\text{м/с}^{2};\ 1\,\text{м/с}^{2}\right)
  3. \left(120\,\text{Н};\ 210\,\text{Н};\ 0\,\text{м/с}^{2};\ 5\,\text{м/с}^{2}\right)
  4. \left(12\,\text{Н};\ 21\,\text{Н};\ 0\,\text{м/с}^{2};\ 0{,}5\,\text{м/с}^{2}\right)
Show the walkthroughHide the walkthrough
Given
m
6\,\text{кг}
\mu
0{,}2
a
1{,}5\,\tfrac{\text{м}}{\text{с}^{2}}
F'
5\,\text{Н}
F''
15\,\text{Н}
Find

F_1F_2a'a''

  1. F_{\text{тр}} = \mu m gF_{\text{тр}} = 0{,}2\cdot 6\cdot 10F_1 = 12\,\text{Н}
  2. F_1 = F_{\text{тр}}F_1 = 12\,\text{Н}

    For uniform movement, traction balances friction.

  3. F_2 = F_{\text{тр}} + maF_2 = 12 + 6\cdot 1{,}5F_2 = 21\,\text{Н}
  4. F' = 5\,\text{Н} < F_{\text{тр}}a' = 0

    The body will not move: a static friction force of 5 N arises.

  5. a'' = \dfrac{F'' - F_{\text{тр}}}{m}a'' = \dfrac{15 - 12}{6}a'' = 0{,}5\,\tfrac{\text{м}}{\text{с}^{2}}

AnswerF_1 = 12\,\text{Н}F_2 = 21\,\text{Н}a' = 0a'' = 0{,}5\,\tfrac{\text{м}}{\text{с}^{2}}

Position in this set: 8 from 32

Problem #42

Dynamics: forces and balance

Masses of weights are suspended from the ends of a weightless lever 4\,\text{кг} and 24\,\text{кг}. Distance from fulcrum to smaller load 60\,\text{см}. Determine the length of the lever.

Answer options

  1. 70\,\text{см}
  2. 35\,\text{см}
  3. 140\,\text{см}
  4. 700\,\text{см}
Show the walkthroughHide the walkthrough
Given
m_1
4\,\text{кг}
m_2
24\,\text{кг}
\ell_1
60\,\text{см}
Find

\ell

  1. m_1 g\,\ell_1 = m_2 g\,\ell_2

    Rule of moments about the fulcrum.

  2. \ell_2 = \dfrac{m_1\ell_1}{m_2}\ell_2 = \dfrac{4\cdot 60}{24}\ell_2 = 10\,\text{см}
  3. \ell = \ell_1 + \ell_2\ell = 60 + 10\ell = 70\,\text{см}

Answer\ell = 70\,\text{см}

Position in this set: 9 from 32

Problem #43

Dynamics: forces and balance

Homogeneous beam of mass 600\,\text{кг} lies on two supports. One support is located at the end of the beam, the other at a distance of a quarter of the length from the other end. With what force does the beam press on the supports?

Answer options

  1. \left(20000\,\text{Н};\ 40000\,\text{Н}\right)
  2. \left(2000\,\text{Н};\ 4000\,\text{Н}\right)
  3. \left(1000\,\text{Н};\ 2000\,\text{Н}\right)
  4. \left(4000\,\text{Н};\ 8000\,\text{Н}\right)
Show the walkthroughHide the walkthrough
Given
m
600\,\text{кг}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

F_{\text{д}1}F_{\text{д}2}

  1. mg\cdot\dfrac{L}{2} = N_2\cdot\dfrac{3L}{4}

    We take the moments relative to the first support; the center of gravity is in the middle of the beam.

  2. N_2 = \dfrac{mg\cdot 4}{2\cdot 3} = \dfrac{2mg}{3}N_2 = \dfrac{2\cdot 600\cdot 10}{3}F_{\text{д}2} = 4000\,\text{Н}
  3. N_1 + N_2 = mg\ \Rightarrow\ N_1 = mg - N_2N_1 = 6000 - 4000F_{\text{д}1} = 2000\,\text{Н}
  4. F_{\text{д}1} = N_1,\quad F_{\text{д}2} = N_2F_{\text{д}1} = 2000\,\text{Н}\ \text{и}\ 4000\,\text{Н}

    According to Newton's third law, the beam presses on the supports with the same forces.

AnswerF_{\text{д}1} = 2000\,\text{Н}F_{\text{д}2} = 4000\,\text{Н}

Position in this set: 10 from 32

Problem #44

Dynamics: Newton's laws

Man of mass 80\,\text{кг} presses on the floor of the elevator with force 700\,\text{Н}. Determine the magnitude and direction of acceleration of the elevator.

Answer options

  1. 0{,}8\,\text{м/с}^{2}\ \text{вниз}
  2. 0\,\text{м/с}^{2}
  3. 1{,}25\,\text{м/с}^{2}\ \text{вниз}
  4. 1{,}25\,\text{м/с}^{2}\ \text{вверх}
Show the walkthroughHide the walkthrough
Given
m
80\,\text{кг}
F_{\text{д}}
700\,\text{Н}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

a

  1. F_{\text{д}} < mg\quad\left(700 < 800\right)\vec{a}\ \text{направлено вниз}

    A person presses less than he weighs, which means the elevator accelerates down or slows down when ascending.

  2. N = F_{\text{д}} = 700\,\text{Н}

    According to Newton's third law.

  3. mg - N = ma\ \Rightarrow\ a = \dfrac{mg - N}{m}a = \dfrac{80\cdot 10 - 700}{80}a = 1{,}25\,\tfrac{\text{м}}{\text{с}^{2}}

Answera = 1{,}25\,\tfrac{\text{м}}{\text{с}^{2}}\ \text{вниз}

Position in this set: 11 from 32

Problem #45

Dynamics: forces and balance

Under the influence of horizontal force 12\,\text{Н} block of mass 5\,\text{кг} moving with acceleration 2\,\text{м/с}^{2}. Determine the coefficient of friction.

Answer options

  1. 0{,}02\,\text{безразм.}
  2. 0{,}08\,\text{безразм.}
  3. 0{,}4\,\text{безразм.}
  4. 0{,}04\,\text{безразм.}
Show the walkthroughHide the walkthrough
Given
F
12\,\text{Н}
m
5\,\text{кг}
a
2\,\tfrac{\text{м}}{\text{с}^{2}}
Find

\mu

  1. F - F_{\text{тр}} = ma\ \Rightarrow\ F_{\text{тр}} = F - maF_{\text{тр}} = 12 - 5\cdot 2F_{\text{тр}} = 2\,\text{Н}
  2. F_{\text{тр}} = \mu m g\ \Rightarrow\ \mu = \dfrac{F_{\text{тр}}}{mg}\mu = \dfrac{2}{5\cdot 10}\mu = 0{,}04

    In one line: mu = (F - ma)/(mg).

Answer\mu = 0{,}04

Position in this set: 12 from 32

Problem #46

Dynamics: forces and balance

Body mass 5\,\text{кг} falls from a height 20\,\text{м} for 2{,}1\,\text{с}. Determine the force of air resistance. Accept g=9{,}81\,\text{м/с}^{2}. The body is released without initial velocity; In this model, consider the resistance force to be constant.

Answer options

  1. 3{,}7\,\text{Н}
  2. 1{,}85\,\text{Н}
  3. 7{,}4\,\text{Н}
  4. 37\,\text{Н}
Show the walkthroughHide the walkthrough
Given
m
5\,\text{кг}
h
20\,\text{м}
t
2{,}1\,\text{с}
g
9{,}81\,\tfrac{\text{м}}{\text{с}^{2}}
Find

F_{\text{с}}

  1. h = \dfrac{at^{2}}{2}\ \Rightarrow\ a = \dfrac{2h}{t^{2}}a = \dfrac{2\cdot 20}{2{,}1^{2}}a = 9{,}07\,\tfrac{\text{м}}{\text{с}^{2}}

    The actual acceleration is less than g, which means that the air is slowing down the body.

  2. mg - F_{\text{с}} = ma\ \Rightarrow\ F_{\text{с}} = m\left(g - a\right)5\left(9{,}81 - 9{,}07\right)F_{\text{с}} = 3{,}7\,\text{Н}

AnswerF_{\text{с}} = 3{,}7\,\text{Н}

Position in this set: 13 from 32

Problem #47

Dynamics: Newton's laws

Force 6\,\text{Н} imparts acceleration to the body 0{,}4\,\text{м/с}^{2}. What force will impart acceleration to this body? 5\,\text{м/с}^{2}?

Answer options

  1. 750\,\text{Н}
  2. 75\,\text{Н}
  3. 37{,}5\,\text{Н}
  4. 150\,\text{Н}
Show the walkthroughHide the walkthrough
Given
F_1
6\,\text{Н}
a_1
0{,}4\,\tfrac{\text{м}}{\text{с}^{2}}
a_2
5\,\tfrac{\text{м}}{\text{с}^{2}}
Find

F_2

  1. \dfrac{F_1}{F_2} = \dfrac{a_1}{a_2}\ \Rightarrow\ F_2 = \dfrac{F_1 a_2}{a_1}F_2 = \dfrac{6\cdot 5}{0{,}4}F_2 = 75\,\text{Н}

    The mass of the body is the same, so the force is proportional to the acceleration.

AnswerF_2 = 75\,\text{Н}

Position in this set: 14 from 32

Problem #48

Dynamics: forces and balance

Body mass 2\,\text{кг} slides on a horizontal surface. A horizontal force acts along its velocity 5\,\text{Н}. The sliding friction coefficient is equal to 0{,}4. Find the projection of acceleration onto the direction of velocity before the body stops.

Answer options

  1. -3\,\text{м/с}^{2}
  2. -15\,\text{м/с}^{2}
  3. -1{,}5\,\text{м/с}^{2}
  4. -0{,}75\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
m
2\,\text{кг}
F
5\,\text{Н}
\mu
0{,}4
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

a_x

  1. F - F_{\text{тр}} = ma_x\ \Rightarrow\ a_x = \dfrac{F - \mu m g}{m}\dfrac{5 - 0{,}4\cdot 2\cdot 10}{2} = \dfrac{5 - 8}{2}a_x = -1{,}5\,\tfrac{\text{м}}{\text{с}^{2}}
  2. F < F_{\text{тр}}\ \Rightarrow\ a_x < 0

    The object is already sliding. The net force opposes its velocity, so it slows to a stop. The negative acceleration component describes this slowing down.

Answera_x = -1{,}5\,\tfrac{\text{м}}{\text{с}^{2}}

Position in this set: 15 from 32

Problem #49

Dynamics: forces and balance

What should be the coefficient of friction so that a motorcyclist can pass a turn with a radius 50\,\text{м} at speed 36\,\text{км/ч}?

Answer options

  1. 0{,}1\,\text{безразм.}
  2. 0{,}4\,\text{безразм.}
  3. 2\,\text{безразм.}
  4. 0{,}2\,\text{безразм.}
Show the walkthroughHide the walkthrough
Given
R
50\,\text{м}
v
36\,\text{км/ч}10\,\text{м/с}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

\mu

  1. F_{\text{тр}} = \dfrac{mv^{2}}{R}

    During a turn, friction plays the role of centripetal force.

  2. \mu m g = \dfrac{mv^{2}}{R}\ \Rightarrow\ \mu = \dfrac{v^{2}}{gR}\mu = \dfrac{10^{2}}{10\cdot 50}\mu = 0{,}2

    The mass is reduced - the result does not depend on it.

Answer\mu = 0{,}2

Position in this set: 16 from 32

Problem #50

Dynamics: Newton's laws

Man of mass 50\,\text{кг} presses on the floor of the elevator with force 600\,\text{Н}. Determine the projection of the elevator's acceleration onto the vertical axis directed upward.

Answer options

  1. 2\,\text{м/с}^{2}
  2. 1\,\text{м/с}^{2}
  3. 4\,\text{м/с}^{2}
  4. 20\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
m
50\,\text{кг}
F_{\text{д}}
600\,\text{Н}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

a

  1. F_{\text{д}} > mg\quad\left(600 > 500\right)\vec{a}\ \text{направлено вверх}
  2. N = F_{\text{д}};\quad N - mg = ma\ \Rightarrow\ a = \dfrac{N - mg}{m}a = \dfrac{600 - 50\cdot 10}{50}a = 2\,\tfrac{\text{м}}{\text{с}^{2}}

Answera = 2\,\tfrac{\text{м}}{\text{с}^{2}}

Position in this set: 17 from 32

Problem #51

Dynamics: forces and balance

At what speed should a cyclist travel so that at the top point of the bridge the radius of curvature 45\,\text{м} push on the bridge with a force half that of gravity?

Answer options

  1. 150\,\text{м/с}
  2. 15\,\text{м/с}
  3. 7{,}5\,\text{м/с}
  4. 30\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
R
45\,\text{м}
F_{\text{д}}
\dfrac{mg}{2}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

v

  1. N = F_{\text{д}} = \dfrac{mg}{2}

    At the top point of the bridge, the force of gravity and the reaction of the support are in opposite directions.

  2. mg - N = ma_{\text{цс}},\quad a_{\text{цс}} = \dfrac{v^{2}}{R}
  3. mg - \dfrac{mg}{2} = \dfrac{mv^{2}}{R}\ \Rightarrow\ v = \sqrt{R\left(g - \dfrac{g}{2}\right)}\sqrt{45\left(10 - 5\right)} = \sqrt{225}v = 15\,\tfrac{\text{м}}{\text{с}} = 54\,\tfrac{\text{км}}{\text{ч}}

Answerv = 15\,\tfrac{\text{м}}{\text{с}}

Position in this set: 18 from 32

Problem #52

Dynamics: forces and balance

What is the force holding the car's mass 1200\,\text{кг} at a turn with a radius 250\,\text{м} at speed 90\,\text{км/ч}? What is the maximum speed at the coefficient of friction 0{,}5?

Answer options

  1. \left(6000\,\text{Н};\ 70{,}72\,\text{м/с}\right)
  2. \left(30000\,\text{Н};\ 353{,}6\,\text{м/с}\right)
  3. \left(3000\,\text{Н};\ 35{,}36\,\text{м/с}\right)
  4. \left(1500\,\text{Н};\ 17{,}68\,\text{м/с}\right)
Show the walkthroughHide the walkthrough
Given
m
1200\,\text{кг}
R
250\,\text{м}
v
90\,\text{км/ч}25\,\text{м/с}
\mu
0{,}5
Find

Fv_{\max}

  1. F = \dfrac{mv^{2}}{R}F = \dfrac{1200\cdot 25^{2}}{250}F = 3000\,\text{Н}
  2. \mu m g = \dfrac{mv_{\max}^{2}}{R}\ \Rightarrow\ v_{\max} = \sqrt{\mu g R}\sqrt{0{,}5\cdot 10\cdot 250} = \sqrt{1250}v_{\max} = 35{,}4\,\tfrac{\text{м}}{\text{с}}

AnswerF = 3000\,\text{Н}v_{\max} = 35{,}4\,\tfrac{\text{м}}{\text{с}}

Position in this set: 19 from 32

Problem #53

Dynamics: Newton's laws

The elevator is moving with acceleration 2\,\text{м/с}^{2}. The weight presses on the floor with force 2880\,\text{Н}. What two masses of a load are possible depending on the direction of acceleration?

Answer options

  1. \left(120\,\text{кг};\ 180\,\text{кг}\right)
  2. \left(480\,\text{кг};\ 720\,\text{кг}\right)
  3. \left(2400\,\text{кг};\ 3600\,\text{кг}\right)
  4. \left(240\,\text{кг};\ 360\,\text{кг}\right)
Show the walkthroughHide the walkthrough
Given
a
2\,\tfrac{\text{м}}{\text{с}^{2}}
F_{\text{д}}
2880\,\text{Н}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

m_1m_2

  1. N = F_{\text{д}};\quad N - mg = ma_x\ \Rightarrow\ N = m\left(g + a_x\right)

    The sign of acceleration depends on where the elevator is accelerating.

  2. m_1 = \dfrac{N}{g + a}m_1 = \dfrac{2880}{10 + 2}m_1 = 240\,\text{кг}

    Acceleration is directed upward - overload, more weight.

  3. m_2 = \dfrac{N}{g - a}m_2 = \dfrac{2880}{10 - 2}m_2 = 360\,\text{кг}

    Acceleration is directed downwards - partial weightlessness.

Answerm_1 = 240\,\text{кг}m_2 = 360\,\text{кг}

Position in this set: 20 from 32

Problem #62

Dynamics: forces and balance

Ball mass 500\,\text{г} hanging on a long thread 80\,\text{см}. He is taken to 90^{\circ} and let go. Determine the thread tension at the lowest point.

Answer options

  1. 15\,\text{Н}
  2. 7{,}5\,\text{Н}
  3. 30\,\text{Н}
  4. 150\,\text{Н}
Show the walkthroughHide the walkthrough
Given
m
500\,\text{г}0{,}5\,\text{кг}
R = \ell
80\,\text{см}0{,}8\,\text{м}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

T

  1. mgh = \dfrac{mv^{2}}{2}\ \Rightarrow\ v = \sqrt{2gh}v = \sqrt{2\cdot 10\cdot 0{,}8}v = 4\,\tfrac{\text{м}}{\text{с}}

    The ball is retracted 90°, so it descends exactly the length of the thread.

  2. T - mg = \dfrac{mv^{2}}{R}\ \Rightarrow\ T = m\left(g + \dfrac{v^{2}}{R}\right)0{,}5\left(10 + \dfrac{4^{2}}{0{,}8}\right)T = 15\,\text{Н}

AnswerT = 15\,\text{Н}

Position in this set: 21 from 32

Problem #147

Newton's laws

Resultant force on a body of mass 4 kg is equal to 12 H. Find the acceleration.

Answer options

  1. 3\,\text{м/с}^{2}
  2. 6\,\text{м/с}^{2}
  3. 1{,}5\,\text{м/с}^{2}
  4. 9\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
m
4\,\text{кг}
F
12\,\text{Н}
Find

a

  1. a=\frac{F}{m}a=\frac{12}{4}a=3\,\text{м/с}^{2}

    We divide the resultant force, not the individual force, by the mass.

Answer3\,\text{м/с}^{2}

Position in this set: 22 from 32

Problem #148

Newton's laws

Resultant force on a body of mass 5 kg is equal to 20 H. Find the acceleration.

Answer options

  1. 8\,\text{м/с}^{2}
  2. 4\,\text{м/с}^{2}
  3. 2\,\text{м/с}^{2}
  4. 12\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
m
5\,\text{кг}
F
20\,\text{Н}
Find

a

  1. a=\frac{F}{m}a=\frac{20}{5}a=4\,\text{м/с}^{2}

    We divide the resultant force, not the individual force, by the mass.

Answer4\,\text{м/с}^{2}

Position in this set: 23 from 32

Problem #149

Newton's laws

Body mass 3 kg is pulled to the right by force 20 N, the resistance force to the left is 5 H. Find the acceleration module.

Answer options

  1. 10\,\text{м/с}^{2}
  2. 5\,\text{м/с}^{2}
  3. 2{,}5\,\text{м/с}^{2}
  4. 15\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
m
3\,\text{кг}
F_1
20\,\text{Н}
F_2
5\,\text{Н}
Find

a

  1. F=F_1-F_2F=20 - 5F=15\,\text{Н}

    We choose the positive direction to the right.

  2. a=\frac{F}{m}a=\frac{15}{3}a=5\,\text{м/с}^{2}

    We apply Newton's law to the resultant.

Answer5\,\text{м/с}^{2}

Position in this set: 24 from 32

Problem #150

Newton's laws

Body mass 4 kg is pulled to the right by force 30 N, the resistance force to the left is 6 H. Find the acceleration module.

Answer options

  1. 12\,\text{м/с}^{2}
  2. 3\,\text{м/с}^{2}
  3. 6\,\text{м/с}^{2}
  4. 18\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
m
4\,\text{кг}
F_1
30\,\text{Н}
F_2
6\,\text{Н}
Find

a

  1. F=F_1-F_2F=30 - 6F=24\,\text{Н}

    We choose the positive direction to the right.

  2. a=\frac{F}{m}a=\frac{24}{4}a=6\,\text{м/с}^{2}

    We apply Newton's law to the resultant.

Answer6\,\text{м/с}^{2}

Position in this set: 25 from 32

Problem #151

Newton's laws

Body mass 2 kg was at rest. A constant horizontal force was applied to it 14 N at constant resistance force 4 N. Find a way behind 4 s.

Answer options

  1. 80\,\text{м}
  2. 20\,\text{м}
  3. 40\,\text{м}
  4. 120\,\text{м}
Show the walkthroughHide the walkthrough
Given
m
2\,\text{кг}
F
14\,\text{Н}
R
4\,\text{Н}
t
4\,\text{с}
Find

S

  1. F_{\text{рез}}=F-RF_{\text{рез}}=14 - 4F_{\text{рез}}=10\,\text{Н}

    Subtract the opposing force.

  2. a=\frac{F_{\text{рез}}}{m}a=\frac{10}{2}a=5\,\text{м/с}^{2}

    Mass relates force and acceleration.

  3. S=\frac{at^2}{2}S=\frac{5 \cdot {4}^{2}}{2}S=40\,\text{м}

    The initial speed is zero, the acceleration is constant.

Answer40\,\text{м}

Position in this set: 26 from 32

Problem #152

Newton's laws

Body mass 3 kg was at rest. A constant horizontal force was applied to it 21 N at constant resistance force 9 N. Find a way behind 5 s.

Answer options

  1. 100\,\text{м}
  2. 25\,\text{м}
  3. 150\,\text{м}
  4. 50\,\text{м}
Show the walkthroughHide the walkthrough
Given
m
3\,\text{кг}
F
21\,\text{Н}
R
9\,\text{Н}
t
5\,\text{с}
Find

S

  1. F_{\text{рез}}=F-RF_{\text{рез}}=21 - 9F_{\text{рез}}=12\,\text{Н}

    Subtract the opposing force.

  2. a=\frac{F_{\text{рез}}}{m}a=\frac{12}{3}a=4\,\text{м/с}^{2}

    Mass relates force and acceleration.

  3. S=\frac{at^2}{2}S=\frac{4 \cdot {5}^{2}}{2}S=50\,\text{м}

    The initial speed is zero, the acceleration is constant.

Answer50\,\text{м}

Position in this set: 27 from 32

Problem #153

Forces and equilibrium

Body mass 4 kg slides horizontally. Friction coefficient 0{,}2, g = 10 m/s². Find the friction force if there are no other vertical forces.

Answer options

  1. 16\,\text{Н}
  2. 8\,\text{Н}
  3. 4\,\text{Н}
  4. 24\,\text{Н}
Show the walkthroughHide the walkthrough
Given
m
4\,\text{кг}
u
0{,}2\,\text{безразм.}
g
10\,\text{м/с}^{2}
Find

F_{\text{тр}}

  1. F_{\text{тр}}=\mu mgF_{\text{тр}}=0{,}2 \cdot 4 \cdot 10F_{\text{тр}}=8\,\text{Н}

    The normal reaction is mg.

Answer8\,\text{Н}

Position in this set: 28 from 32

Problem #154

Forces and equilibrium

Body mass 6 kg slides horizontally. Friction coefficient 0{,}3, g = 10 m/s². Find the friction force if there are no other vertical forces.

Answer options

  1. 36\,\text{Н}
  2. 9\,\text{Н}
  3. 18\,\text{Н}
  4. 54\,\text{Н}
Show the walkthroughHide the walkthrough
Given
m
6\,\text{кг}
u
0{,}3\,\text{безразм.}
g
10\,\text{м/с}^{2}
Find

F_{\text{тр}}

  1. F_{\text{тр}}=\mu mgF_{\text{тр}}=0{,}3 \cdot 6 \cdot 10F_{\text{тр}}=18\,\text{Н}

    The normal reaction is mg.

Answer18\,\text{Н}

Position in this set: 29 from 32

Problem #155

Forces and equilibrium

The elevator accelerates up with acceleration 2 m/s². Passenger weight 60 kg, g = 10 m/s². Find the weight of the passenger.

Answer options

  1. 1440\,\text{Н}
  2. 360\,\text{Н}
  3. 720\,\text{Н}
  4. 2160\,\text{Н}
Show the walkthroughHide the walkthrough
Given
a
2\,\text{м/с}^{2}
m
60\,\text{кг}
g
10\,\text{м/с}^{2}
Find

P

  1. F=maF=60 \cdot 2F=120\,\text{Н}

    The resultant is directed upward.

  2. P=mg+maP=60 \cdot 10 + 120P=720\,\text{Н}

    The support reaction exceeds the force of gravity by ma.

Answer720\,\text{Н}

Position in this set: 30 from 32

Problem #156

Forces and equilibrium

The elevator accelerates up with acceleration 3 m/s². Passenger weight 70 kg, g = 10 m/s². Find the weight of the passenger.

Answer options

  1. 1820\,\text{Н}
  2. 455\,\text{Н}
  3. 2730\,\text{Н}
  4. 910\,\text{Н}
Show the walkthroughHide the walkthrough
Given
a
3\,\text{м/с}^{2}
m
70\,\text{кг}
g
10\,\text{м/с}^{2}
Find

P

  1. F=maF=70 \cdot 3F=210\,\text{Н}

    The resultant is directed upward.

  2. P=mg+maP=70 \cdot 10 + 210P=910\,\text{Н}

    The support reaction exceeds the force of gravity by ma.

Answer910\,\text{Н}

Position in this set: 31 from 32

Problem #157

Forces and equilibrium

Spring stiffness 200 N/m stretched by 0{,}1 m and pulls horizontally on a block of mass 2 kg, sliding in the direction of the spring force. The friction coefficient is 0{,}2, g = 10 m/s². Find the acceleration at this moment.

Answer options

  1. 16\,\text{м/с}^{2}
  2. 4\,\text{м/с}^{2}
  3. 24\,\text{м/с}^{2}
  4. 8\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
k
200\,\text{Н/м}
x
0{,}1\,\text{м}
m
2\,\text{кг}
u
0{,}2\,\text{безразм.}
g
10\,\text{м/с}^{2}
Find

a

  1. F=kxF=200 \cdot 0{,}1F=20\,\text{Н}

    The elastic force at the moment under consideration.

  2. R=\mu mgR=0{,}2 \cdot 2 \cdot 10R=4\,\text{Н}

    Friction is directed against sliding.

  3. a=\frac{F-R}{m}a=\frac{(20 - 4)}{2}a=8\,\text{м/с}^{2}

    We find the instantaneous acceleration using the resultant.

Answer8\,\text{м/с}^{2}

Position in this set: 32 from 32

Problem #158

Forces and equilibrium

Spring stiffness 300 N/m stretched by 0{,}12 m and pulls horizontally on a block of mass 3 kg, sliding in the direction of the spring force. The friction coefficient is 0{,}2, g = 10 m/s². Find the acceleration at this moment.

Answer options

  1. 10\,\text{м/с}^{2}
  2. 20\,\text{м/с}^{2}
  3. 5\,\text{м/с}^{2}
  4. 30\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
k
300\,\text{Н/м}
x
0{,}12\,\text{м}
m
3\,\text{кг}
u
0{,}2\,\text{безразм.}
g
10\,\text{м/с}^{2}
Find

a

  1. F=kxF=300 \cdot 0{,}12F=36\,\text{Н}

    The elastic force at the moment under consideration.

  2. R=\mu mgR=0{,}2 \cdot 3 \cdot 10R=6\,\text{Н}

    Friction is directed against sliding.

  3. a=\frac{F-R}{m}a=\frac{(36 - 6)}{3}a=10\,\text{м/с}^{2}

    We find the instantaneous acceleration using the resultant.

Answer10\,\text{м/с}^{2}

The theory behind these problems

Work through the topic if a problem will not give in.

Free fall and circular motionFall without resistance and the kinematics of uniform rotation.Newton's lawsHow forces relate to acceleration and interactions between bodies.Forces and equilibriumGravity, elasticity, friction, inclined plane and moment of force.

Formulas by topic

Look up symbols and units of measurement in the reference.

Apparent weight (force on a support)P · НSpring constantk · Н/мCoefficient of frictionμ · безразм.Torque (moment of a force)M · Н·мPeriodT · сDistance and pathS, l · мForceF · НAngular velocityω · рад/сExtensionΔl · мFrequencyν · Гц