Work, power and energy

Two smooth slopes start at the same height and end at the bottom. One is steeper, the other longer. If identical bodies are released from rest, which will be faster at the bottom? We can answer without calculating acceleration along every section.

When we lift a load, an interaction transfers energy through motion. Work done by a force describes this transfer. What matters is the force component along the displacement of its point of application: a forward push does positive work, a braking force negative work. A perpendicular component can turn the velocity of a point particle without changing its magnitude. A large force with no displacement does not, by itself, imply mechanical work.

For a constant force, its component along the displacement is F\cos\alpha. Thus the work is A=FS\cos\alpha, where F is the force magnitude, S is the displacement magnitude of its point of application, and \alpha is the angle between them. At 0° the work is greatest and positive; at 90° it is zero; at 180° it is negative. If the force varies along the path, divide the motion into small sections and add their work: one force value may not describe the whole path.

To connect work to measurement, take a force of 1 N and a displacement of 1 m in its direction. The work is 1 J: 1\,\text{Дж}=1\,\text{Н}\cdot\text{м}=1\,\text{кг}\cdot\text{м}^2/\text{с}^2. Work and energy share a unit because work describes energy transfer. Turn the same force perpendicular to the displacement and its work becomes zero, even though the force magnitude and distance travelled are unchanged.

Accelerate a body along a straight line using a constant force. Dynamics gives F=ma and kinematics gives v^2-v_0^2=2aS. Multiply the second relation by m/2 to obtain FS=mv^2/2-mv_0^2/2: work changes the quantity E_{\text{к}}=mv^2/2. This is kinetic energy. The formula describes translational motion in classical mechanics; a rotating body also has rotational energy. Kilograms and metres per second give joules.

Doubling mass at the same speed doubles kinetic energy; doubling speed makes it four times larger. Stopping a fast-moving body therefore requires much more negative work. With the same braking force, four times the energy requires four times the stopping distance. Direction does not change the sign of energy: two bodies moving towards each other can have zero total momentum and positive total kinetic energy.

A raised load can speed up as it falls. The energy associated with the relative positions of the body and Earth is potential energy, E_{\text{п}}=mgh, in a uniform field. Height h is measured from a chosen zero level. Move that level upwards: all values change, but the difference between two positions stays the same. Negative potential energy is therefore allowed; observable energy exchange depends on its change, not on the arbitrary zero.

During a descent from h_1 to h_2, gravity does work A_{\text{тяж}}=mg(h_1-h_2). Every downward vertical step contributes positively, every upward step negatively, and horizontal steps contribute nothing. Details of the path cancel, leaving only the height difference. If the start and finish are at the same height, gravity does zero total work. This explains why a long smooth slope can give the same final speed as a short steep one.

When stretching a spring, the force is not constant: it grows from zero to k\Delta l. The area under the force graph is a triangle, giving stored energy E_{\text{упр}}=k(\Delta l)^2/2. Doubling the deformation quadruples the energy because the average force doubles too. Equal magnitudes of compression and extension store equal energy. Choose zero energy at the undeformed spring; Hooke's law must hold throughout the range considered.

Let the spring deformation change from x_1 to x_2. As the spring returns towards its original shape, stored energy decreases and transfers into other forms; the elastic force does work A_{\text{упр}}=kx_1^2/2-kx_2^2/2. When the deformation magnitude decreases, this work is positive. If we stretch the spring further, its force opposes displacement and does negative work. Equal initial and final deformation magnitudes give zero work.

Two motors lift the same load through the same height, but one does it faster. The work is equal; the rate of energy transfer differs. Power N describes that rate: average power is N_{\text{ср}}=A/\Delta t, with unit 1\,\text{Вт}=1\,\text{Дж/с}. For the same work, half the time requires twice the average power. Here N means power, whereas in the previous lesson it denoted the normal force; context and units distinguish them.

In a short time, the point of application of the force moves v\Delta t. Dividing the work by that time gives instantaneous power N=\vec F\cdot\vec v=Fv\cos\alpha. With force and velocity in the same direction, doubling either doubles power if the other stays fixed. When they are perpendicular, power is zero. A load held still receives no mechanical energy this way, even though the person's muscles consume internal energy.

On a smooth slope, energy of position decreases while energy of motion increases. Combine them into mechanical energy E_0=E_{\text{к}}+E_{\text{п}}. The potential part includes interactions of the selected system, such as a body with Earth and a spring. This is energy accounting: we must not count the same transfer both as work done by gravity and as an extra loss in addition to the change in its potential energy.

If forces not included in potential energy do zero total work, then E_{\text{к}1}+E_{\text{п}1}=E_{\text{к}2}+E_{\text{п}2}. On a stationary smooth slope, the normal force is perpendicular to motion and does no work. Equal initial speeds and height differences therefore give equal final speeds. Descent times may differ: the energy balance answers a speed question but does not, by itself, give the time.

If the remaining work is nonzero, the balance for the body and the chosen conservative interactions is \Delta E_0=A_{\text{ост}}. A driving force can increase mechanical energy; friction against a stationary surface can reduce it. The lost part does not vanish: internal energy of the body and surface increases, and sound may be produced. Deforming and warming parts require a full energy balance; mechanical energy is not all the energy.

A 2 kg body slides from rest without friction from a height of 5 m. Find its speed at the bottom, taking g=10 м/с².

The system includes the body and Earth. There is no friction, and the normal force is perpendicular to motion, so it does no work.

Choose zero potential energy at the bottom. Initially E= mgh = 2×10×5 = 100 Дж.

At the bottom, the same energy is kinetic: mv²/2=100. With m=2 кг, we obtain v²=100.

v=10 м/с. Mass would cancel from the general equation: the result does not depend on the body's mass.

The speed at the bottom is 10 m/s. Initially, the system had 100 J of energy of position; at the bottom, the body has 100 J of energy of motion. A heavier body would start with more energy, but would need proportionally more energy to reach the same speed.

Work accounts for the force component along the motion. Newton's law and kinematics give the energy of motion mv²/2.

Potential energy describes stored energy of an interaction; its change has meaning regardless of the chosen zero.

Smooth slopes with the same height difference give the same final speed for the same initial conditions. Friction requires us to include conversion into internal energy.

Next, we will use forces to understand pressure in a stationary liquid.