Question

A ball of mass 2.00 kg is dropped from a height of 1.50 m (from the ground) onto a massless spring (the spring has an equilibrium length of 0.500 m). The ball compresses the spring by an amount of 0.200 m by the time it comes to a stop. Calculate the spring constant of the spring.

A. 250 N/m
B. 500 N/m
C. 625 N/m
D. 1000 N/m

Answer 1

The spring constant of the spring is C. 625 N/m. So, Option C is correct.

Step-by-step explanation:

When the ball drops and compresses the spring, the potential energy gained by the ball due to the height it was dropped from is converted into the potential energy stored in the spring during compression. The potential energy (PE) can be equated to the elastic potential energy stored in the spring using the formula:

`\[ PE = (1)/(2) k x^2 \]`

where k is the spring constant and x is the compression of the spring. The potential energy gained by the ball is given by mgh , where m is the mass, g is the acceleration due to gravity, and h is the height.

Setting these two expressions for potential energy equal to each other, we get:

`\[ mgh = (1)/(2) k x^2 \]`

Rearranging for k , we find:

`\[ k = (2mgh)/(x^2) \]`

Substituting the given values m = 2.00 kg, g = 9.8 m/s² , h = 1.50 m, and x = 0.200 m, we get:

`\[ k = \frac{2 * 2.00 \, \text{kg} * 9.8 \, \text{m/s}^2 * 1.50 \, \text{m}}{(0.200 \, \text{m})^2} \]`

Solving this expression gives k = 625 N/m, which corresponds to option C. Therefore, the spring constant of the spring is 625 N/m.