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Energy loss due to friction

by fernancb
Tags: energy loss, friction, work
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fernancb
#1
Feb22-11, 04:51 PM
P: 28
1. The problem statement, all variables and given/known data
A 15.0 kg block is dragged over a rough, horizontal surface by a 74.9 N force acting at 17.6 degrees above the horizontal. The block is displaced 4.91 m, and the coefficient of kinetic friction is 0.266.

Find the work done by the 74.9 N force = 351 J
Find the work done by the normal force = 0 J
What work does the gravitational force do on the block? = 0 J

This is the part i can't figure out:

How much energy is lost due to friction?

And related to that:
Find the total change in the block's kinetic energy.

2. Relevant equations
Ff = ((mu)k)(mg)
E = F*d


3. The attempt at a solution
Ef = ((mu)k)(mg) (d)
= 0.266*15*9.8*4.91
= 192 J

What am I doing wrong?
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PhanthomJay
#2
Feb22-11, 07:36 PM
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The friction force is not (mu_k)mg, it is (mu_k)N, where N is the Normal force, which , in this problem, is not equal to mg. Solve for the normal force first. The use the work energy equation or other method to get the change in KE.
fernancb
#3
Feb22-11, 07:41 PM
P: 28
So it'll be : Ff = (0.266)(74.9 sin17.6)
= 6.02 N

fernancb
#4
Feb22-11, 07:47 PM
P: 28
Energy loss due to friction

Then I took that and put it in to W = Fd = (6.02)(4.91) = 29.6 J

Still wrong...?
SammyS
#5
Feb22-11, 07:47 PM
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Quote Quote by fernancb View Post
So it'll be : Ff = (0.266)(74.9 sin17.6)
= 6.02 N
The normal force is not 74.9 sin(17.6) either.

Draw a Free Body Diagram.
fernancb
#6
Feb22-11, 07:49 PM
P: 28
Huh? do i use the mass in there somewhere?
PhanthomJay
#7
Feb22-11, 08:45 PM
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When you draw a free body diagram, you note all forces acting on the block , in both the x and y directions. There are three forces acting in the y direction, one of which is the component of the applied force which you have correctly calculated. What are the other 2 forces acting on the block in the y direction? Then use one of newton's laws to find the unknown force in that direction.
fernancb
#8
Feb24-11, 08:46 AM
P: 28
So the frictional force is:
Ff = (mu)k * N
= (0.266)* (74.9sin17.6 + (15*9.8))
= 45.1 N

But how would I calculate the work done? The block doesn't move in the y-direction, so I would think no work coule be done since W=Fd
PhanthomJay
#9
Feb24-11, 09:00 AM
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Quote Quote by fernancb View Post
So the frictional force is:
Ff = (mu)k * N
= (0.266)* (74.9sin17.6 + (15*9.8))
= 45.1 N
You are not handling the plus/minus signs corerctly when calculating the normal force. The normal force and vert comp of the applied force act up, and the weight force acts down. The algebraic sum of these 3 forces adds up to 0, per application of Newton 1 in the y direction.

But how would I calculate the work done? The block doesn't move in the y-direction, so I would think no work coule be done since W=Fd
In the y direction, yes, there is no work done. But there is work done in the x direction. Find the work done in the x direction by the friction force. That is the energy lost due to friction. Then use energy methods to calculate the kinetic energy change.
fernancb
#10
Feb24-11, 09:40 AM
P: 28
Okay, so my normal force would then be

Fg = N + vertF
N= Fg - vertF
= (15*9.8) - (74.9sin17.6)
= 124 N

then my frictional force would be:
Ff = (mu)k * N
= (0.266) * (124N)
= 33.0 N

Work done by friction:
W = Fd
= 33N * 4.91m
= 162 J
PhanthomJay
#11
Feb24-11, 11:28 AM
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Quote Quote by fernancb View Post
Okay, so my normal force would then be

Fg = N + vertF
N= Fg - vertF
= (15*9.8) - (74.9sin17.6)
= 124 N

then my frictional force would be:
Ff = (mu)k * N
= (0.266) * (124N)
= 33.0 N

Work done by friction:
W = Fd
= 33N * 4.91m
= 162 J
Be careful with signs...the work done by friction is -162 J , since the friction force is opposite the direction of the block's motion. The energy lost due to friction is 162 J. Now solve for the change in the object's kinetic energy, and watch signage, please.


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