Question

Consider a frictionless ramp angled 20° above the horizontal w/an uncompressed spring at the bottom. The spring constant of this spring is 30N/m . If a 19-N box slides 2.8 m down the ramp, then hits the spring and compresses it 72cm from its equilibrium position, whats the net work done on the box?

Answer 1

We know that work done is the change in energy. That is it the difference between initial total energy and final total energy.

Let the initial height be H1. The box travels from a height H1 to a distance of 2.8 m downwards and compresses the spring by a distance x = 72 cm = 0.72 m. After compressing the spring, the box is at a height H2 from the ground. Now we taking the height H2 as the reference line to solve the problem.

Taking the height H2 as reference line, let the total distance travelled by the box be y. Now

y = 2.8 + x = 2.8 + 0.72 = 3.52 m

Now the work done by the box is

W = Final total energy - Initial total energy

W = (Final Potential energy + Final Kinetic Energy) - (Initial Potential energy + Initial Kinetic energy)

W = ((PE)final + (KE)final) - ((PE)initial + (KE)initial)

The total energy of the box is converted as the potential energy of the spring when it compresses So final potential energy is the potential energy of the spring.

Since the box comes to rest (KE)final = 0

Initial potential energy of the box, (PE)initial = mg (H1 - H2). We get H1 - H2 = y Sin , where is the angle of inclination. Therefore

(PE)initial = mg (y Sin )

Since the box starts from rest, (KE)initial = 0

Given = 20 degree, spring constant (k) = 30 N/m and weight of the box = 19 N.

Weight = mg = 19 N

Taking g = 9.8 m/s²

Substituting the values in the equation of work

W = 7.78 - 22.89 = -12.11 J

Net work done on the box is -12.11 J