Q: Suppose the spring in the first image has a spring constant $$k = 200\ \mathrm{N/m}$$ and is compressed by $$0.10\ \mathrm{m}. $$What is the work done by the spring as it returns to its equilibrium position?

$$W = \frac{1}{2}k x^2$ = \frac{1}{2} \times 200 \times (0.10)^2$ = 1\ \mathrm{J}$$

Q: If the block in the second image has a mass of $$0.5\ \mathrm{kg}$$ and the spring is compressed by $$0.20\ \mathrm{m}$$ with $$k = 200\ \mathrm{N/m}$$, what is the speed of the block as it leaves the spring (assuming no friction)?

$$v = \sqrt{\frac{kx^2$}{m}} = \sqrt{\frac{200 \times (0.20)^2$}{0.5}} = 2.53\ \mathrm{m/s}$$

Question 1

Consider the two spring setups shown in the images above. In both cases, a spring is compressed and then released. In the first image, the spring is compressed by a small amount, while in the second image, the spring is compressed by a larger amount. If the same block is launched by each spring, which statement best describes the kinetic energy of the block as it leaves the spring in each case?

Accepted Answer

The block in the second image will have greater kinetic energy because the work done by the spring is proportional to the square of the compression distance: $$W = \frac{1}{2}kx^2$.

Question 2

If the spring in the second image is compressed twice as far as in the first image, by what factor does the kinetic energy of the block change as it leaves the spring?

Accepted Answer

The kinetic energy increases by a factor of 4, since $$K = \frac{1}{2}k(2x)^2$ = 2^2$ 	imes \frac{1}{2}kx^2$ = 4 	imes \frac{1}{2}kx^2$.

Question 3

Suppose the spring in the first image has a spring constant $$k = 200\ \mathrm{N/m}$$ and is compressed by $$0.10\ \mathrm{m}. $$What is the work done by the spring as it returns to its equilibrium position?

Accepted Answer

$$W = \frac{1}{2}k x^2$ = \frac{1}{2} 	imes 200 	imes (0.10)^2$ = 1\ \mathrm{J}$$

Question 4

If the block in the second image has a mass of $$0.5\ \mathrm{kg}$$ and the spring is compressed by $$0.20\ \mathrm{m}$$ with $$k = 200\ \mathrm{N/m}$$, what is the speed of the block as it leaves the spring (assuming no friction)?

Accepted Answer

$$v = \sqrt{\frac{kx^2$}{m}} = \sqrt{\frac{200 	imes (0.20)^2$}{0.5}} = 2.53\ \mathrm{m/s}$$

Work and Kinetic Energy: Concepts, Calculations, and Applications

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