Question

A 3.05-kg ball of clay is thrown downward from a height of 2.75 m with a speed of 5.01 m/s onto a spring with k = 1750 N/m. The clay compresses the spring a certain maximum amount before momentarily stopping. What is the total work done on the clay during the spring's compression? Also, how would one go about solving for that?

Answer 1

The ball will gain kintetic energy as it is moving downwards. KE = 1/2 mass*velocity^2

a(t) = -9.81 m/s

v(t) = -9.81*t - 5.01

x(t) = (1/2)*-9.81*t^2 - 5.01t + 2.75

The time that the ball hits the spring is equal to:

x(t) = 0 = (1/2)*-9.81*t^2 - 5.01t + 2.75. Solve for t. (Quadratic equation)

t = 0.39565 seconds

Plugging this value of t into the velocity equation will give us the velocity of the ball when it makes contact with the spring.

v(0.39565) = -9.81*(0.39565) - 5.01 = -8.8913 m/s

All of the kinetic energy of the ball will be transfered into the spring, so the work that the spring does on the ball is equal to the kintetic energy of the ball at the moment right before it makes contact.

W = KE = 1/2 * m * v^2 = 1/2 * 3.05* (-8.8913)^2 = 120.5599 Joules

An alternative way to look at this problem.

Instead of using position, velocity, and acceleration functions, you can calculate the initial energy of the ball. The initial energy of the ball is the sum of the initial potential energy and the initial kinetic energy. All of the initial energy of the ball will be transfered into the spring, and therefore is equal to the work that the spring does on the ball. The only force acting on the ball as it moves towards the spring is gravity, which is conservative, and therefore does not change the energy of the ball. (If, for example, air resistance was acting on the ball, you would have to account for this as it would remove some energy from the ball before it made contact from the spring)

Potential energy = m * g * h = 3.05 * 9.81 * 2.75 = 82.2777 Joules

Kinetic energy = (1/2) * m * v^2 = 1/2 * 3.05* (-5.01)^2 = 38.2814 Joules

82.2777 + 38.2814 = 120.5590 Joules