Question

A 310 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.2 N/cm (see the figure). The block becomes attached to the spring and compresses the spring 16 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

Answer 1

Below is an example of how to do this problem with slightly different values...that way you can do the problem yourself.

Let's say k=2.2 N/cm or 220 N/m and m=310 grams and the distance was 16 cm.

Gravity has constant force of 9.8*m = 9.8*.31 = 3.038 Newtons. This is for a distance of .16 meters, so that is .16*3.038 = .48608 Newtons

The force on an ideal, Hooke's Law spring is kx, where k is the spring constant and x the distance from the equilibrium point.
The work is found by integrating this to get 1/2 kx^2.
This means the work done is 1/2*220*(.16)^2= 2.816 Joules. [Negative because the force is in the opposite direction as the motion.]

The total work represents the change in kinetic energy. The total work done is -5.022 + .423 = -4.598 Joules.

Equating this to the kinetic energy lost, we can find velocity:

K.E.= 1/2 * .24 * v^2 --> v = square root of 38.322 = 6.2 m/s.

To solve the last part...doubling the speed quadruples the energy, so the block will have an energy of 4.598 * 4 = 18.392 Joules.

If you let X be the distance traveled, the work done by gravity will be .24*9.8*X and the work done by the spring will be 1/2*310*X^2 [to convert to newtons].

This gives 18.392 = -2.352X + 155 X^2

Solving this with the quadratic formula yields X = .35 meters, or 35 centimeters.

This makes sense because the work done by gravity is so small that doubling the speed should almost double the compression