Question

A spring is kept compressed by a toy car of mass $150g$ . On releasing the car, it moves with a speed of $0.2m{s^{ - 1}}$ . So, the elastic potential energy of the spring is
A) $3{\text{ }}mJ$
B) $3{\text{ }}J$
C) $1.5\,{\text{mJ}}$
D) $4{\text{ }}mJ$

Answer 1

In this solution, we will use the law of conservation of energy. The potential of the spring when it is compressed will be completely transferred to the kinetic energy of the car when the spring becomes uncompressed.

Formula used: In this solution, we will use the following formula:
-Kinetic energy of the car: $K = \dfrac{1}{2}m{v^2}$ where $m$ is the mass of the car and $v$ is the velocity of the car.

Complete step by step answer
When the spring is compressed, it will have a potential energy corresponding to the distance it is compressed which is denoted by $U$ .
Now when the car is released, the spring starts uncompressing and in the process converts its potential energy into the kinetic energy of the car. In this process, all of the potential energy will be converted completely into the kinetic energy of the car.
We know that the kinetic energy of a moving object such as a car can be calculated as
$K = \dfrac{1}{2}m{v^2}$
Substituting the value of $m = 150g = 0.15\,kg$ and $v = 0.2\,m/s$ , we get the kinetic energy as
$K = \dfrac{1}{2}0.15{(0.2)^2}$
$\Rightarrow K = 3 \times {10^{ - 3}}\,J$
Or alternatively
$K = 3\,mJ$
Hence the kinetic energy of the car will be 3 mJ which corresponds to option (A).

Note
While calculating the kinetic energy of the car, we must be careful to take all the units in the SI system to calculate the energy in Joules. Here we have ignored frictional forces which can lead to energy losses in the spring car system and prevent the complete conversion of the potential energy of the spring into the kinetic energy of the car.