Question

A spring is kept compressed by a toy cart of
mass of 150g and on releasing the cart it moves with speed of 0.2m/s.....what is potential energy of spring?

Answer 1

Given, m=150g=150×10−3kg,v=0.2ms−1

On releasing the toy car, the spring tries to come back to its uncompressed state by pushing the car away. Thus the potential energy stored in the spring changes into the kinetic energy of the car.

Elastic potential energy of spring(U)=kinetic energy gained by the cart

So, U=1/2×m×v2=1/2×150×10−3×0.22=3×10−3J=3mJ

Answer 2

Answer:

Potential energy of spring is 3 mJ.

Explanation:

Given,A spring is kept compressed by a toy cart of mass of 150g and on releasing the cart it moves with speed of 0.2m/s

So,$m = 150g \: = 150 \times {10}^{ - 3} \: kg \: \:$and $v = 0.2 \: m {s}^{ - 1}$

Here we want find potential energy of spring.

On releasing the toy car, the spring tries to come back to its uncompressed state by pushing the car away. Thus the potential energy stored in the spring changes into the kinetic energy of the car.

Elastic potential energy of spring(U)=kinetic energy gained by the cart.

So,

$U = \frac{1}{2} \times m \times {v}^{2} \\ = \frac{1}{2} \times 150 \times {10}^{ - 3} \times {0.2}^{2} \\ = 3 \times {10}^{ - 3} J \\ = 3 \: mj$