How Does a Spring's Work Relate to Frictional Forces in Motion?

[gagga5]

A relaxed spring with spring constant k = 60 N/m is stretched a distance di = 59 cm and held there. A block of mass M = 7 kg is attached to the spring. The spring is then released from rest and contracts, dragging the block across a rough horizontal floor until it stops without passing through the relaxed position, at which point the spring is stretched by an amount df = di/9.

Ok, so there is three questions.

a) In moving from the initial to the final position, by how much has the kinetic energy of the block changed?

- I found out this is zero, since v = 0, for initial and final.

(b) What is the work done by the spring?

- This question is driving me insane. I know that potential energy equation for the spring is U = 1/2 k(Sf-Si)^2.
So I tried with 1/2*60*(59)^2 - 1/2*60*(59-59/9)^2 = 103141
it was not the answer. also i tried with negative sign, I tried adding them, tried all kinds of possibilities, but all of them were not the answer. I think I should use different way to solve this problem.

(c) What is the magnitude of the total work done by the frictional force?
I guess I can figure this out once i know about the work done by the spring...

(d) What is the magnitude of the frictional force on the block?
This one could probably be solved once I know about c..

So my main question is how do I figure out the work done by the spring.

 

[Doc Al]

(b) What is the work done by the spring?

- This question is driving me insane. I know that potential energy equation for the spring is U = 1/2 k(Sf-Si)^2.

At any point, the amount of PE stored in the spring is given by U = 1/2 k(x)^2, where x is the amount that the spring is stretched from its relaxed position. In this case, the change in spring PE would be 1/2 k (Sf^2 - Si^2).

 

[kyle8921]

To calculate the work done by the spring, we need to use the equation W = ∫F dx, where F is the force applied by the spring and dx is the displacement of the block. In this case, the force applied by the spring can be calculated using Hooke's Law, F = -kx, where k is the spring constant and x is the displacement of the block from its equilibrium position.

In this problem, the block is initially held at a distance of 59 cm from the equilibrium position, and then released to move towards the equilibrium position. However, it does not reach the equilibrium position and stops at a distance of 59/9 cm from the equilibrium position. This means that the total displacement of the block is 59/9 cm - 59 cm = -59/9 cm.

Using this value of displacement, we can now calculate the work done by the spring as W = ∫F dx = ∫(-kx) dx = -k∫x dx = -kx^2/2. Plugging in the values of k and x, we get W = -60*(-59/9)^2/2 = 1170 J.

Therefore, the work done by the spring in this scenario is 1170 J. To calculate the work done by the frictional force, we need to use the equation W = Ff*d, where Ff is the frictional force and d is the displacement of the block. Since the block stops at a distance of 59/9 cm from the equilibrium position, the displacement is 59/9 cm. Plugging in this value and the mass of the block (7 kg) into the equation, we get W = Ff*d = (7 kg)(9.8 m/s^2)(59/9 cm) = 41.2 J.

Therefore, the magnitude of the total work done by the frictional force is 41.2 J. To calculate the magnitude of the frictional force, we can rearrange the equation to Ff = W/d = (41.2 J)/(59/9 cm) = 0.696 N.

In summary, the work done by the spring is 1170 J, the total work done by the frictional force is 41.2 J, and the magnitude of the frictional force is 0.696 N.