Question

(a) Explain by an example what is meant by potential energy. Write down the expression for the gravitational potential energy of a body of mass m placed at height h above the surface of the earth.
(b) What is the difference between potential energy and kinetic energy?
(c) A ball of mass 0.5 kg slows down from a speed of 5 m/s to that of 3 m/s.
Calculate the change in kinetic energy of the ball. State your answer giving proper units.

Answer 1

Hint:We know that every object possesses an energy whether it is moving or in the rest position. The gravitational potential energy is the product of the weight of the body and height of the body above the ground. Recall the difference between kinetic energy and potential energy.

Formula used:
Gravitational potential energy, \[U = mgh\],
where, m is the mass, g is the acceleration due to gravity and h is the height.
Kinetic energy,\[K = \dfrac{1}{2}m{v^2}\],
where v is the velocity.

Complete step by step answer:
(a) We know that every object possesses energy whether it is moving or in the rest position. The energy possessed by the body in its rest position or due to change in the position is referred to as potential energy of the body. When we displace the spring from its mean position, the energy gets stored in the spring in the form of potential energy.The expression for the gravitational potential energy of the body of mass m placed at height h above the ground is,
\[U = mgh\]
Here, g is the acceleration due to gravity.

(b) Let’s discuss the difference between kinetic energy and potential energy. The energy possessed by the body when it is moving with a certain velocity is known as kinetic energy while the energy possessed by the body due to its position is known as potential energy. In kinematics, if the kinetic energy increases, the potential energy decreases.

(c) We have the expression for the kinetic energy of the body,
\[K = \dfrac{1}{2}m{v^2}\]
Let’s determine the change in the kinetic energy of the ball as follows,
\[\Delta K = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2\]
\[ \Rightarrow \Delta K = \dfrac{1}{2}m\left( {v_f^2 - v_i^2} \right)\]
Here, \[{v_f}\] is the final velocity and \[{v_i}\] is the initial velocity.
Substituting 5 m/s for \[{v_i}\] and 3 m/s for \[{v_f}\] in the above equation, we get,
\[\Delta K = \dfrac{1}{2}\left( {0.5} \right)\left( {{3^2} - {5^2}} \right)\]
\[ \therefore \Delta K = - 4\,{\text{J}}\]
The negative sign for the kinetic energy implies that the kinetic energy of the ball is decreased.

Note: If the total energy of the system does not change in the path of the particle, the change in kinetic energy of the particle is equal to negative change in the potential energy of the particle. This is known as conservation of total energy. Remember, the unit of both kinetic energy and potential energy is joule.