Work and Kinetic Energy Theorem

[holmeskaei]

Homework Statement

A 4.5 kg box slides down a 5.1m high frictionless hill, starting from rest, across a 2.4m wide horizontal surface, then hits a horizontal spring with spring constant 540 N/m. The other end of the spring is anchored against a wall. The ground under the spring is frictionless, but the 2.4m long horizontal surface is rough. The coefficient of kinetic friction of the box on this surface is 0.26.

A. What is the speed of the box just before reaching the rough surface?
B. What is the speed of the box just before hitting the spring?
C. How far is the spring compressed?
D. Including the first crossing, how many complete trips will the box make across the rough surface before coming to rest?

Homework Equations

W=(delta)K
K=1/2mv^2
Wnet=sum of all work
Fk=uk(n)(delta)(x)

The Attempt at a Solution

A. I did 1/2mv^2=mg(delta)(x) and I got 1/2(4.5)v=(4.5)(5.1) v=9.10m/s and it is incorrect.
Since I did this incorrectly and don't know why, I couldn't begin to attempt the rest of the problem. Any help is appreciated, thanks!

 

[LowlyPion]

A. I did 1/2mv^2=mg(delta)(x) and I got 1/2(4.5)v=(4.5)(5.1) v=9.10m/s and it is incorrect.
Since I did this incorrectly and don't know why, I couldn't begin to attempt the rest of the problem. Any help is appreciated, thanks!

I'd redo your calculation. I get a different number for the values using g = 9.8.

1/2*m*v² = m*g*h

v² = 2*g*h = 2*9.8*5.1

 

[nlightner]

The Work and Kinetic Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. In this scenario, the box has gravitational potential energy at the top of the hill, which is converted to kinetic energy as it slides down the hill. Once it reaches the rough surface, some of its kinetic energy will be converted to work against friction, causing its speed to decrease.

A. To find the speed of the box just before reaching the rough surface, we can use the Work and Kinetic Energy Theorem. The work done by gravity equals the change in kinetic energy, so we have:

W = (delta)K
mgh = 1/2mv^2

Where m is the mass of the box, g is the gravitational acceleration, h is the height of the hill, and v is the speed of the box. Plugging in the given values, we get:

(4.5 kg)(9.8 m/s^2)(5.1 m) = 1/2(4.5 kg)v^2
v = 7.82 m/s

B. To find the speed of the box just before hitting the spring, we can use the Work and Kinetic Energy Theorem again. The work done by gravity equals the change in kinetic energy, and the work done by friction also equals the change in kinetic energy. So we have:

W = (delta)K
mgh - Fk(delta)x = 1/2mv^2

Where Fk is the force of kinetic friction, and (delta)x is the distance the box travels on the rough surface. Plugging in the given values, we get:

(4.5 kg)(9.8 m/s^2)(5.1 m) - (0.26)(4.5 kg)(9.8 m/s^2)(2.4 m) = 1/2(4.5 kg)v^2
v = 6.09 m/s

C. To find how far the spring is compressed, we can use the formula for potential energy stored in a spring:

Ep = 1/2kx^2

Where k is the spring constant and x is the distance the spring is compressed. We know that at the point of maximum compression, all of the box's kinetic energy will be converted to potential energy stored in the spring. So we can set the two equations equal