An EASY work and energy problem. Is my answer correct?

  • Thread starter DLxX
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I need to know if my answer is correct for the following problem.

A 2.04 kg block slides down a ramp from a height of 4.77m. If friction does 20J or work, what is the speed at the bottom of the ramp?

I keep getting 9.7m/s. Is this correct?
 

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  • #2
learningphysics
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DLxX said:
I need to know if my answer is correct for the following problem.

A 2.04 kg block slides down a ramp from a height of 4.77m. If friction does 20J or work, what is the speed at the bottom of the ramp?

I keep getting 9.7m/s. Is this correct?
I'm not getting that. Can you show what you did?
 
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learningphysics said:
I'm not getting that. Can you show what you did?
I found the work with the W=Fd formula and then used that value for work in the formula W=Ekf - Eki (kinetic energy) (.5mv^2).
 
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dextercioby
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What would the KE be,if it hadn't been friction on the inclined plane...?

Daniel.
 
  • #5
learningphysics
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DLxX said:
I found the work with the W=Fd formula and then used that value for work in the formula W=Ekf - Eki (kinetic energy) (.5mv^2).
Ah, you didn't take friction into account.

You need to include the work done by gravity as well as work done by friction on the left hand side. Be careful with signs.

mgh-20 = Ekf - Eki
 
  • #6
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dextercioby said:
What would the KE be,if it hadn't been friction on the inclined plane...?

Daniel.
95.4? I found that using the Gravitation Potential Energy Formula since I dont know the velocity yet, and since the grav pot should be equal to the kin pot at the end. Right?
 
  • #7
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learningphysics said:
Ah, you didn't take friction into account.

You need to include the work done by gravity as well as work done by friction on the left hand side. Be careful with signs.

mgh-20 = Ekf - Eki
I did that and got the correct answer of 8.6m/s, but would one of you mind explaining the theory behind subtracting the force of friction from the Grav Pot Energy? Should I just think of it like Net force or something?
 
  • #8
learningphysics
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DLxX said:
I did that and got the correct answer of 8.6m/s, but would one of you mind explaining the theory behind subtracting the force of friction from the Grav Pot Energy? Should I just think of it like Net force or something?
To get total work done, you can get work done by each force... you don't know the frictional force... but you're given that it does -20J of work. Only other force left is gravity which does mgh of work (force of mg through a distance of h). So total work is mgh-20.

Two ways to think about this problem come to mind to me. First way is just total work on object = change in kinetic energy... here we count all the forces acting on the object including gravity. This is how you did the problem.

Second way is when we use gravitational potential energy. I'll call mechanical energy= K.E + G.P.E

MEfinal - MEinitial = work done by all forces except gravity.

Both ways give the same answer. You can verify this.
 
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  • #9
learningphysics
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Work done by gravity through a distance h is mgh. (force=mg distance=h). Just wanted to point out that mgh here is not being used as GPE but as work done by gravity.

The math comes out the same, but it gives two slightly different ways of thinking of the problem.
 

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