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GRADE 7 PRACTICE

Physics Problems: Forces, Grade 7

Gravity, elasticity, weight, resultant and friction.

27 problems across 5 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.

27 problems
Position in this set: 1 from 27

Problem #25

Force, gravity and apparent weight

What concept does the description correspond to: “A vector quantity that measures interaction between objects and can change motion or deform an object.»?

Answer options

  1. \text{Force}
  2. \text{Gravitational force}
  3. \text{Apparent weight and weightlessness}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{A vector quantity that measures interaction between objects and can change motion or deform an object.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    A force has a magnitude, direction and point of application. The newton is the SI unit of force.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Force}

Answer\text{Concept} = \text{Force}

Position in this set: 2 from 27

Problem #26

Force, gravity and apparent weight

What concept does the description correspond to: “The attractive force exerted on an object by a celestial body; near the surface it points vertically towards that body's centre.»?

Answer options

  1. \text{Gravitational force}
  2. \text{Force}
  3. \text{Apparent weight and weightlessness}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{The attractive force exerted on an object by a celestial body; near the surface it points vertically towards that body's centre.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    On another planet, the object's mass stays the same, but gravitational force changes with the free-fall acceleration.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Gravitational force}

Answer\text{Concept} = \text{Gravitational force}

Position in this set: 3 from 27

Problem #27

Gravitational force

A mass of 4 kg is located at the surface of the Earth. Take the acceleration due to gravity to be 9.8 m/s². Find the force of gravity.

Answer options

  1. 39{,}2\,\text{Н}
  2. 2{,}45\,\text{Н}
  3. 13{,}8\,\text{Н}
  4. 98\,\text{Н}
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Given
m
4\,\text{кг}
g
9{,}8\,\text{м/с}^{2}
Find

F_{\text{тяж}}

  1. F_{\text{тяж}}=mgF_{\text{тяж}}=4\cdot9{,}8\,\text{Н}F_{\text{тяж}} = 39{,}2\,\text{Н}

Answer39{,}2\,\text{Н}

Position in this set: 4 from 27

Problem #28

Elasticity and net force

What concept does the description correspond to: “A change in an object's shape or size under external forces is called deformation.»?

Answer options

  1. \text{Deformation}
  2. \text{Hooke's law}
  3. \text{Net force}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{A change in an object's shape or size under external forces is called deformation.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Elastic deformation disappears when the load is removed; plastic deformation leaves a noticeable permanent change.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Deformation}

Answer\text{Concept} = \text{Deformation}

Position in this set: 5 from 27

Problem #29

Elasticity and net force

What concept does the description correspond to: “In the linear elastic range, the magnitude of the restoring force is proportional to the magnitude of the change in length.»?

Answer options

  1. \text{Hooke's law}
  2. \text{Deformation}
  3. \text{Net force}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{In the linear elastic range, the magnitude of the restoring force is proportional to the magnitude of the change in length.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    The spring constant relates force to change in length. Recovering the original shape does not guarantee linearity: elastic deformation can extend beyond the range of Hooke's law.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Hooke's law}

Answer\text{Concept} = \text{Hooke's law}

Position in this set: 6 from 27

Problem #30

Hooke's law

A spring with a stiffness of 200 N/m is stretched by 3 cm. Find the modulus of elasticity.

Answer options

  1. 6\,\text{Н}
  2. 0{,}06\,\text{Н}
  3. 66{,}7\,\text{Н}
  4. 600\,\text{Н}
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Given
k
200\,\text{Н/м}
\Delta l
3\,\text{см}0{,}03\,\text{м}
Find

F_{\text{упр}}

  1. F_{\text{упр}}=k\Delta lF_{\text{упр}}=200\cdot0{,}03\,\text{Н}F_{\text{упр}} = 6\,\text{Н}

Answer6\,\text{Н}

Position in this set: 7 from 27

Problem #31

Friction force

What concept does the description correspond to: “A force that prevents the onset of relative sliding of contacting surfaces.»?

Answer options

  1. \text{Static friction}
  2. \text{Rolling resistance}
  3. \text{Kinetic friction}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{A force that prevents the onset of relative sliding of contacting surfaces.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    It is static friction that allows the foot to push off the ground without slipping and keeps the object on an inclined surface.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Static friction}

Answer\text{Concept} = \text{Static friction}

Position in this set: 8 from 27

Problem #32

Friction force

What concept does the description correspond to: “Resistance to a rolling object's motion, usually smaller than kinetic friction under the same conditions.»?

Answer options

  1. \text{Rolling resistance}
  2. \text{Static friction}
  3. \text{Kinetic friction}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{Resistance to a rolling object's motion, usually smaller than kinetic friction under the same conditions.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Wheels reduce losses during transportation, but friction remains necessary for acceleration, braking and steering.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Rolling resistance}

Answer\text{Concept} = \text{Rolling resistance}

Position in this set: 9 from 27

Problem #33

Friction force

The box slides along the horizontal floor. The coefficient of friction is 0.2, the reaction force of the support is 50 N. Find the friction force.

Answer options

  1. 10\,\text{Н}
  2. 0{,}004\,\text{Н}
  3. 25\,\text{Н}
  4. 250\,\text{Н}
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Given
\mu
0{,}2
N
50\,\text{Н}
Find

F_{\text{тр}}

  1. F_{\text{тр}}=\mu NF_{\text{тр}}=0{,}2\cdot50\,\text{Н}F_{\text{тр}} = 10\,\text{Н}

Answer10\,\text{Н}

Position in this set: 10 from 27

Problem #106

Force, gravity and apparent weight

Object mass 3 kg. Find force gravity at g = 10 N/kg.

Answer options

  1. 30\,\text{Н}
  2. 60\,\text{Н}
  3. 15\,\text{Н}
  4. 90\,\text{Н}
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Given
m
3\,\text{кг}
g
10\,\text{Н/кг}
Find

F

  1. F=mgF=3 \cdot 10F=30\,\text{Н}

    The force of gravity acts on a body from the Earth.

Answer30\,\text{Н}

Position in this set: 11 from 27

Problem #107

Force, gravity and apparent weight

Object mass 7 kg. Find force gravity at g = 10 N/kg.

Answer options

  1. 140\,\text{Н}
  2. 70\,\text{Н}
  3. 35\,\text{Н}
  4. 210\,\text{Н}
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Given
m
7\,\text{кг}
g
10\,\text{Н/кг}
Find

F

  1. F=mgF=7 \cdot 10F=70\,\text{Н}

    The force of gravity acts on a body from the Earth.

Answer70\,\text{Н}

Position in this set: 12 from 27

Problem #108

Force, gravity and apparent weight

motionless body presses on horizontal support with by force 4 N. Find him mass B grams at g = 10 N/kg.

Answer options

  1. 800\,\text{г}
  2. 400\,\text{г}
  3. 200\,\text{г}
  4. 1200\,\text{г}
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Given
P
4\,\text{Н}
g
10\,\text{Н/кг}
Find

M

  1. m=\frac{P}{g}m=\frac{4}{10}m=0{,}4\,\text{кг}

    At rest, the weight is equal in magnitude to the force of gravity.

  2. M=1000mM=0{,}4 \cdot 1000M=400\,\text{г}

    Convert kilograms to grams.

Answer400\,\text{г}

Position in this set: 13 from 27

Problem #109

Force, gravity and apparent weight

motionless body presses on horizontal support with by force 7{,}5 N. Find him mass B grams at g = 10 N/kg.

Answer options

  1. 1500\,\text{г}
  2. 375\,\text{г}
  3. 750\,\text{г}
  4. 2250\,\text{г}
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Given
P
7{,}5\,\text{Н}
g
10\,\text{Н/кг}
Find

M

  1. m=\frac{P}{g}m=\frac{7{,}5}{10}m=0{,}75\,\text{кг}

    At rest, the weight is equal in magnitude to the force of gravity.

  2. M=1000mM=0{,}75 \cdot 1000M=750\,\text{г}

    Convert kilograms to grams.

Answer750\,\text{г}

Position in this set: 14 from 27

Problem #110

Force, gravity and apparent weight

Empty vessel has mass 0{,}5 kg. B him poured 3 l water density 1 kg/l, then cast 1 l. Find weight vessel with remaining water. g = 10 N/kg, vessel motionless.

Answer options

  1. 50\,\text{Н}
  2. 12{,}5\,\text{Н}
  3. 25\,\text{Н}
  4. 75\,\text{Н}
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Given
m_0
0{,}5\,\text{кг}
V
3\,\text{л}
r
1\,\text{кг/л}
v
1\,\text{л}
g
10\,\text{Н/кг}
Find

P

  1. m=\rho(V-v)m=1 \cdot (3 - 1)m=2\,\text{кг}

    Find the mass of the remaining water.

  2. M=m_0+mM=0{,}5 + 2M=2{,}5\,\text{кг}

    We take into account the mass of the vessel.

  3. P=MgP=2{,}5 \cdot 10P=25\,\text{Н}

    At rest the weight of the system is Mg.

Answer25\,\text{Н}

Position in this set: 15 from 27

Problem #111

Force, gravity and apparent weight

Empty vessel has mass 0{,}8 kg. B him poured 5 l water density 1 kg/l, then cast 2 l. Find weight vessel with remaining water. g = 10 N/kg, vessel motionless.

Answer options

  1. 76\,\text{Н}
  2. 19\,\text{Н}
  3. 114\,\text{Н}
  4. 38\,\text{Н}
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Given
m_0
0{,}8\,\text{кг}
V
5\,\text{л}
r
1\,\text{кг/л}
v
2\,\text{л}
g
10\,\text{Н/кг}
Find

P

  1. m=\rho(V-v)m=1 \cdot (5 - 2)m=3\,\text{кг}

    Find the mass of the remaining water.

  2. M=m_0+mM=0{,}8 + 3M=3{,}8\,\text{кг}

    We take into account the mass of the vessel.

  3. P=MgP=3{,}8 \cdot 10P=38\,\text{Н}

    At rest the weight of the system is Mg.

Answer38\,\text{Н}

Position in this set: 16 from 27

Problem #112

Elasticity and net force

Spring stiffness 200 N/m stretched by 0{,}03 m B within elastic deformation. Find force elasticity.

Answer options

  1. 12\,\text{Н}
  2. 6\,\text{Н}
  3. 3\,\text{Н}
  4. 18\,\text{Н}
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Given
k
200\,\text{Н/м}
x
0{,}03\,\text{м}
Find

F

  1. F=kxF=200 \cdot 0{,}03F=6\,\text{Н}

    We apply Hooke's law to the magnitude of the force.

Answer6\,\text{Н}

Position in this set: 17 from 27

Problem #113

Elasticity and net force

Spring stiffness 300 N/m stretched by 0{,}04 m B within elastic deformation. Find force elasticity.

Answer options

  1. 24\,\text{Н}
  2. 6\,\text{Н}
  3. 12\,\text{Н}
  4. 36\,\text{Н}
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Given
k
300\,\text{Н/м}
x
0{,}04\,\text{м}
Find

F

  1. F=kxF=300 \cdot 0{,}04F=12\,\text{Н}

    We apply Hooke's law to the magnitude of the force.

Answer12\,\text{Н}

Position in this set: 18 from 27

Problem #114

Elasticity and net force

Force 8 N stretches Spring on 4 cm. Find her Hardness B N/m.

Answer options

  1. 400\,\text{Н/м}
  2. 100\,\text{Н/м}
  3. 200\,\text{Н/м}
  4. 600\,\text{Н/м}
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Given
F
8\,\text{Н}
x
4\,\text{см}
Find

k

  1. l=\frac{x}{100}l=\frac{4}{100}l=0{,}04\,\text{м}

    We convert the elongation into meters.

  2. k=\frac{F}{l}k=\frac{8}{0{,}04}k=200\,\text{Н/м}

    Stiffness is equal to the ratio of force to elongation.

Answer200\,\text{Н/м}

Position in this set: 19 from 27

Problem #115

Elasticity and net force

Force 15 N stretches Spring on 5 cm. Find her Hardness B N/m.

Answer options

  1. 600\,\text{Н/м}
  2. 150\,\text{Н/м}
  3. 900\,\text{Н/м}
  4. 300\,\text{Н/м}
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Given
F
15\,\text{Н}
x
5\,\text{см}
Find

k

  1. l=\frac{x}{100}l=\frac{5}{100}l=0{,}05\,\text{м}

    We convert the elongation into meters.

  2. k=\frac{F}{l}k=\frac{15}{0{,}05}k=300\,\text{Н/м}

    Stiffness is equal to the ratio of force to elongation.

Answer300\,\text{Н/м}

Position in this set: 20 from 27

Problem #116

Elasticity and net force

On ring act right strength 12 and 8 N, left strength 5 N. Ring holds Spring stiffness 300 N/m, pulling left. Find Elongation springs B centimeters.

Answer options

  1. 10\,\text{см}
  2. 2{,}5\,\text{см}
  3. 15\,\text{см}
  4. 5\,\text{см}
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Given
F_1
12\,\text{Н}
F_2
8\,\text{Н}
F_3
5\,\text{Н}
k
300\,\text{Н/м}
Find

l

  1. F=F_1+F_2-F_3F=12 + 8 - 5F=15\,\text{Н}

    The spring force balances the other forces.

  2. x=\frac{F}{k}x=\frac{15}{300}x=0{,}05\,\text{м}

    We find the elongation using Hooke's law.

  3. l=100xl=0{,}05 \cdot 100l=5\,\text{см}

    Convert the result to centimeters.

Answer5\,\text{см}

Position in this set: 21 from 27

Problem #117

Elasticity and net force

On ring act right strength 20 and 10 N, left strength 6 N. Ring holds Spring stiffness 400 N/m, pulling left. Find Elongation springs B centimeters.

Answer options

  1. 6\,\text{см}
  2. 12\,\text{см}
  3. 3\,\text{см}
  4. 18\,\text{см}
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Given
F_1
20\,\text{Н}
F_2
10\,\text{Н}
F_3
6\,\text{Н}
k
400\,\text{Н/м}
Find

l

  1. F=F_1+F_2-F_3F=20 + 10 - 6F=24\,\text{Н}

    The spring force balances the other forces.

  2. x=\frac{F}{k}x=\frac{24}{400}x=0{,}06\,\text{м}

    We find the elongation using Hooke's law.

  3. l=100xl=0{,}06 \cdot 100l=6\,\text{см}

    Convert the result to centimeters.

Answer6\,\text{см}

Position in this set: 22 from 27

Problem #118

Friction force

block evenly slides right by table. His pull right by force 8 N, A Additional strength 2 N directed left. Find force friction.

Answer options

  1. 12\,\text{Н}
  2. 3\,\text{Н}
  3. 6\,\text{Н}
  4. 18\,\text{Н}
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Given
F
8\,\text{Н}
P
2\,\text{Н}
Find

F_{\text{тр}}

  1. F_{\text{тр}}=F-PF_{\text{тр}}=8 - 2F_{\text{тр}}=6\,\text{Н}

    With uniform linear motion, the horizontal forces are balanced.

Answer6\,\text{Н}

Position in this set: 23 from 27

Problem #119

Friction force

block evenly slides right by table. His pull right by force 12 N, A Additional strength 3 N directed left. Find force friction.

Answer options

  1. 18\,\text{Н}
  2. 4{,}5\,\text{Н}
  3. 27\,\text{Н}
  4. 9\,\text{Н}
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Given
F
12\,\text{Н}
P
3\,\text{Н}
Find

F_{\text{тр}}

  1. F_{\text{тр}}=F-PF_{\text{тр}}=12 - 3F_{\text{тр}}=9\,\text{Н}

    With uniform linear motion, the horizontal forces are balanced.

Answer9\,\text{Н}

Position in this set: 24 from 27

Problem #120

Friction force

B experience Kinetic friction force amounts to 0{,}2 from strength pressing. On block valid strength pressing 30 N. Find force friction.

Answer options

  1. 12\,\text{Н}
  2. 3\,\text{Н}
  3. 18\,\text{Н}
  4. 6\,\text{Н}
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Given
r
0{,}2\,\text{доля}
N
30\,\text{Н}
Find

F_{\text{тр}}

  1. F_{\text{тр}}=rNF_{\text{тр}}=0{,}2 \cdot 30F_{\text{тр}}=6\,\text{Н}

    We use the force ratio given in the condition.

Answer6\,\text{Н}

Position in this set: 25 from 27

Problem #121

Friction force

B experience Kinetic friction force amounts to 0{,}3 from strength pressing. On block valid strength pressing 40 N. Find force friction.

Answer options

  1. 12\,\text{Н}
  2. 24\,\text{Н}
  3. 6\,\text{Н}
  4. 36\,\text{Н}
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Given
r
0{,}3\,\text{доля}
N
40\,\text{Н}
Find

F_{\text{тр}}

  1. F_{\text{тр}}=rNF_{\text{тр}}=0{,}3 \cdot 40F_{\text{тр}}=12\,\text{Н}

    We use the force ratio given in the condition.

Answer12\,\text{Н}

Position in this set: 26 from 27

Problem #122

Friction force

block of mass 2 kg and cargo on him mass 1 kg evenly pull together by table. Friction force equal to 0{,}2 from strength pressing. Find horizontal force traction at g = 10 N/kg.

Answer options

  1. 6\,\text{Н}
  2. 12\,\text{Н}
  3. 3\,\text{Н}
  4. 18\,\text{Н}
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Given
m_1
2\,\text{кг}
m_2
1\,\text{кг}
r
0{,}2\,\text{доля}
g
10\,\text{Н/кг}
Find

F

  1. m=m_1+m_2m=2 + 1m=3\,\text{кг}

    Both bodies add load to the table.

  2. N=mgN=3 \cdot 10N=30\,\text{Н}

    With horizontal thrust, the pressure is equal to the force of gravity.

  3. F=F_{\text{тр}}=rNF=0{,}2 \cdot 30F=6\,\text{Н}

    Even traction balances friction.

Answer6\,\text{Н}

Position in this set: 27 from 27

Problem #123

Friction force

block of mass 3 kg and cargo on him mass 2 kg evenly pull together by table. Friction force equal to 0{,}3 from strength pressing. Find horizontal force traction at g = 10 N/kg.

Answer options

  1. 30\,\text{Н}
  2. 15\,\text{Н}
  3. 7{,}5\,\text{Н}
  4. 45\,\text{Н}
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Given
m_1
3\,\text{кг}
m_2
2\,\text{кг}
r
0{,}3\,\text{доля}
g
10\,\text{Н/кг}
Find

F

  1. m=m_1+m_2m=3 + 2m=5\,\text{кг}

    Both bodies add load to the table.

  2. N=mgN=5 \cdot 10N=50\,\text{Н}

    With horizontal thrust, the pressure is equal to the force of gravity.

  3. F=F_{\text{тр}}=rNF=0{,}3 \cdot 50F=15\,\text{Н}

    Even traction balances friction.

Answer15\,\text{Н}

The theory behind these problems

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

Force, gravity and apparent weightForce as a measure of interaction, gravitational force, apparent weight and weightlessness.Elasticity and net forceDeformation, Hooke's law, force meters and adding forces along a straight line.Friction forceStatic friction, kinetic friction and rolling resistance in technology and everyday life.

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.ForceF · НExtensionΔl · м