Question

The spring of a toy car is wound by pushing the car

Answer 1

7.5 Joules is the amount of elastic potential energy stored in the car's spring during this process.

w=Fd

w=(15)(0.5)

w = 7.5

Answer 2

The elastic potential energy stored in the car's spring during the process is 3.75 J

Determination of the spring constant

From the question given above, the following data were obtained:

• Force (F) = 15 N

• Extention (e) = 0.5 m

• Spring constant (K) =?

K = F/e

K = 15 / 0.5

K = 30 N/m

Determination of the potential energy

• Spring constant (K) = 30 N/m

• Extention (e) = 0.5 m

• Potential energy (PE) =?

PE = ½Ke²

PE = ½ × 30 × 0.5²

PE = 15 × 0.25

PE = 3.75 J

Therefore, the elastic potential energy stored in the car's spring during the process is 3.75 J

Learn more about energy stored in spring:

brainly.com/question/4280346