Finding work done by friction force

[Agent M27]

Homework Statement

A 19 kg block is dragged over a rough, horizontal surface by a constant force of 174N acting at an angle of 34.9 above the horizontal. The block is displaced 54m, and the coefficient of kinetic friction is .233. The acceleration of gravity is 9.8 ms^-2. Find the work done by the force of friction.

Homework Equations

W=$$\Sigma$$F_xxD

f_k = $$\mu$$n

$$\Sigma$$F_x = Fcos$$\theta$$-f_k

The Attempt at a Solution

So I think that my online homework module is incorrect on this one because I don't see how it could be any other way. Here is what I have done:

W=$$\Sigma$$F_x*d = [174Ncos(34.9)-(.233)(19Kg)(9.8ms^-2)]*54m = 5363.37863J.

Now this is the net-work done but others online show this as the correct way to find the work done by the friction force. I tried it and it was incorrect. Then I re-read the problem and though maybe it was just f_k*d which gave me 2342.7684J which is also incorrect. Where am I going wrong here or am I correct in my assumption that the online module is wrong on this one? Thanks in advance.

Joe

 

[collinsmark]

Hello Joe,

W=$$\Sigma$$F_x*d = [174Ncos(34.9)-(.233)(19Kg)(9.8ms^-2)]*54m = 5363.37863J.

Now this is the net-work done but others online show this as the correct way to find the work done by the friction force. I tried it and it was incorrect. Then I re-read the problem and though maybe it was just f_k*d which gave me 2342.7684J which is also incorrect. Where am I going wrong here or am I correct in my assumption that the online module is wrong on this one? Thanks in advance.

A couple of pointers.

(I) The problem statement asks to find the work done by friction. That might give you a different answer than the work done by whatever was pulling the rope (definitely will be a different answer if the object accelerates). So in general, it makes a difference. Concentrate on the friction.

(II) As an interim step of finding the force of friction, you need to calculate the normal force. As usual, gravity plays a part when calculating the normal force. Are there any other forces (or components of which) that have an effect on the normal force?

 

[Agent M27]

Well being that the block is in vertical equilibrium the normal force ought to equal the force of gravity with no other components to it, that changes if there is an inclined plane though. Using that normal force and multiplying it by the coefficient of friction, that gives me the force of kinetic friction, the work done by that force ought to be that value times the displacement, but it is incorrect. I guess I don't see why that is not the work done by the kinetic friction force. Thanks for the help though.

Joe

 

[nrqed]

Well being that the block is in vertical equilibrium

True

the normal force ought to equal the force of gravity with no other components to it,

Watch out. The net force along the vertical is zero. If you draw a free body diagram you will see that the normal force is not equal to the force of gravity here.

that changes if there is an inclined plane though. Using that normal force and multiplying it by the coefficient of friction, that gives me the force of kinetic friction, the work done by that force ought to be that value times the displacement, but it is incorrect. I guess I don't see why that is not the work done by the kinetic friction force. Thanks for the help though.

Joe

Another hint: the work done by a force is given by $$W = F d cos \theta$$
where d is the distance the object moved (so it is a positive number by definition) and $$\theta$$ is the angle between the direction of the force and the direction of the displacement.


Hope this helps!

 

[Agent M27]

I'm on my cell phone so please excuse lack of Tex. So I went ahead and drew a free body diagram of the box with n and mg being perpendicular and the force coming off the block at an angle. From that I pressume my n will equal mgsin(theta) which I then plug into my kinetic friction force eq. That solution multiplied by the displacement gives me roughly 1340 j, is that correct? Thanks for the assistance.

Joe

 

[collinsmark]

Hello Joe,

I'm on my cell phone so please excuse lack of Tex. So I went ahead and drew a free body diagram of the box with n and mg being perpendicular and the force coming off the block at an angle. From that I pressume my n will equal mgsin(theta) which I then plug into my kinetic friction force eq. That solution multiplied by the displacement gives me roughly 1340 j, is that correct? Thanks for the assistance.

Something is still not quite right . I think it might be the "I pressume my n will equal mgsin(theta)," if by n, you mean the normal force.

The normal force will be a addition or subtraction of two separate vector components. You should expect to see a '+' or a '-' sign in your equation for the normal force.

Also, the block is on a horizontal surface, so there shouldn't be a trigonometric function [such as sin(theta)] associated with the gravitational part.

(Hint: Imagine putting a bathroom scale underneath a block, on a horizontal surface, such that the bathroom scale measures the normal force. Now grab the top of the block and pull up on it. How does that affect the scale's reading? )

 

[Agent M27]

Thanks again Collinsmark, I was thinking about it totally incorrect, I was in a rush. What I have found is that the normal force is the difference of the force along the vertical direction (breaking the force into its components) and the force of gravity. The relationship is as follows:

N=(Fsin$$\theta$$-mg) Which makes sense (to me) because I am pulling upward on the block with some force in the y direction and gravity is counter acting that force keeping it in constant contact with the friction surface.

Which then means the work done by the friction force is as follows W_f=$$\mu$$nd

= -1090.1877 J of work done by the friction force on the block.

Thanks again for the assistance.

Joe