Calculating Net Work

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A 391 kg piano slides 3.7 m down a(n) 27° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 7-23). The effective coefficient of kinetic friction is 0.40.

Figure 7-23

(a) Calculate the force exerted by the man.
374N
(b) Calculate the work done by the man on the piano.
-1384J
(c) Calculate the work done by the friction force.
-5054J
(d) Calculate the work done by the force of gravity.
6438J
(e) Calculate the net work done on the piano.
_____J

I got the previous questions correct with the exception of letter "e".
This is what I did...

WORKnet = Wgravity-Wfriction+Wnormal force in x direction-Work man in x direction

My numbers were 6438-5054+6437-1384 = 6437J
The problem must be in the way I'm calculating the normal force in the x direction. I used Fn=Fg*sin27*3.7m=6437J

Thanks in advance for the help.
 

Answers and Replies

  • #2
Doc Al
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Fanman22 said:
I got the previous questions correct with the exception of letter "e".
This is what I did...

WORKnet = Wgravity-Wfriction+Wnormal force in x direction-Work man in x direction
The normal force has no component in the x-direction (parallel to the incline) and thus does no work.

The problem must be in the way I'm calculating the normal force in the x direction. I used Fn=Fg*sin27*3.7m=6437J
The normal force equals the component of the weight normal to the incline: [itex]F_n = mg cos(27)[/itex]. But, as previously mentioned, it acts perpendicular to the motion and thus does no work.

A shortcut way to get the net work on an object: The net work equals the change in KE.
 
  • #3
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so I should simply be able to omit the WORKnormalforce component and get the answer....but its still wrong. Since there is no velocity given, and there is no accleration or time, then it has a KE of 0, right?

Maybe I can use the change in potential energy? I can find the length of all sides since the 27degree angle is given and the distance 3.7 is given. Find the mgh at the max height, then find the mgh at the min height. Will the difference be the -Work ?
 
  • #4
Doc Al
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1,502
Fanman22 said:
so I should simply be able to omit the WORKnormalforce component and get the answer....but its still wrong.
What makes you say that? Just add the values you found for b, c, and d to find the total work.
Since there is no velocity given, and there is no accleration or time, then it has a KE of 0, right?
The KE depends on the speed, which you don't know. But all you care about is the change in KE, which is zero.

Maybe I can use the change in potential energy?
You've already incorporated gravitational potential energy by caculating the work done by gravity. Don't count it twice.
 

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