Question

The 3.0 kg object below is released from rest at a height of 5.0 m on a curved frictionless ramp. At the foot of the ramp is a spring of force constant k = 500 N/m. The object slides down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. a) Find x. b) what happens to the object after it comes to rest?

Answer 1

Answer:

a) The compressing distance is 0.8 m.

b) After the object comes to rest, it will be released with an initial speed of 7.1 m/s and will reach a height of 5.0 m.

Explanation:

Hi there!

a)Initially, the object has only gravitational potential energy that is calculated as follows:

PE = m· g · h

Where:

PE = potential energy.

m = mass.

g = acceleration due to gravity.

h = height.

Then:

PE = 3.0 kg · 9.8 m/s² · 5.0 m = 1.5 × 10² J

When the object reaches the spring and compresses it, all that potential energy is converted into elastic potential energy (there is no energy dissipation as heat because there is no friction). The elastic potential energy is calculated as follows:

EPE = 1/2 · k · x²

Then, the elastic potential energy is equal to the initial gravitational energy:

PE = EPE = 1.5 × 10² J = 1/2 · 500 N/m · x²

Solving for x:

1.5 × 10² Nm / (1/2 · 500 N /m) = x²

0.6 m² = x²

x = 0.8 m

The compressing distance is 0.8 m.

b) After the object comes to rest, the object will be released with a kinetic energy equal to the elastic potential energy. We can calculate the initial velocity of the object after it is released with the following equation:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass.

v = speed.

Since initially EPE = KE:

1.5 × 10² J = 3.0 kg · v²

√( 1.5 × 10² J / 3.0 kg) = v

v = 7.1 m/s

Then, after the object comes to rest, it will be released with an initial speed of 7.1 m/s and will reach a height of 5.0 m.