Question

A 25kg block is placed 2.7m above the ground on a frictionless ramp. At the bottom of a ramp there is an ideal spring.

A. If the block is let go, how fast will the block be moving at the bottom of the ramp?

B. The block runs into the spring and compresses the spring a total of .35m before coming to a stop. What is the spring constant of the spring?

The block then slides back up the ramp. How high (above the ground) does the block get before coming to a stop?

Answer 1

Mass of block = m = 25 kg

Height of the block above the ground = h = 2.7 m

Speed of the block at the bottom of the ramp = V₁

Spring constant = k

Compression of the spring before the block comes to a stop = X = 0.35 m

Gravitational acceleration = g = 9.81 m/s²

The initial potential energy of the block at the top of the ramp is converted into kinetic energy at the bottom of the ramp.

V₁ = 7.28 m/s

The kinetic energy of the block is converted into the potential energy of the spring when the block comes to rest.

k = 10816 N/m

The potential energy of the spring is converted into kinetic energy of the block again as it is pushed away from the block and the into potential energy as is goes up the ramp.

h_new = 2.7 m

The block will go to the same initial height it was released from as there is no energy loss throughout the motion.

A) Speed of block at the bottom of the ramp = 7.28 m/s

B) Spring constant = 10816 N/m

The block reaches a height of 2.7m before coming to a stop as it slides back up the ramp.