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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?

- Thread starter DLxX
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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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learningphysics

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I'm not getting that. Can you show what you did?DLxX said:

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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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).learningphysics said:I'm not getting that. Can you show what you did?

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What would the KE be,if it hadn't been friction on the inclined plane...?

Daniel.

Daniel.

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learningphysics

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Ah, you didn't take friction into account.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).

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

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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?dextercioby said:What would the KE be,if it hadn't been friction on the inclined plane...?

Daniel.

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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?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

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learningphysics

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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.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?

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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learningphysics

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

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