A mass m = 17 kg is pulled along a horizontal floor, with a coefficient of kinet

Question

A mass m = 17 kg is pulled along a horizontal floor, with a coefficient of kinetic friction ?k = 0.06, for a distance d = 6.6 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle ? = 33 with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of ? = 33 (thus on the incline it is parallel to the surface) and has a tension T = 45 N.

1)

What is the work done by tension before the block gets to the incline?

J  

2)

What is the work done by friction as the block slides on the flat horizontal surface?

J  

3)

What is the speed of the block right before it begins to travel up the incline?

m/s  

4)

How far up the incline does the block travel before coming to rest?

m  

5)

What is the work done by gravity as it comes to rest?

J  

6)

During the entire process, the net work done on the block is:

positive

negative

zero

A mass m = 17 kg is pulled along a horizontal floor, with a coefficient of kinetic friction ?k = 0.06, for a distance d = 6.6 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle ? = 33½ with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of ? = 33½ (thus on the incline it is parallel to the surface) and has a tension T = 45 N. 1) What is the work done by tension before the block gets to the incline? J 2) What is the work done by friction as the block slides on the flat horizontal surface? J 3) What is the speed of the block right before it begins to travel up the incline? m/s 4) How far up the incline does the block travel before coming to rest? m 5) What is the work done by gravity as it comes to rest? J 6) During the entire process, the net work done on the block is: positive negative zero

Explanation / Answer

Work Done By tension W = T * ds

W = F cos theta * ds

W = 45 * 6.6 * cos 33

W = 249 Joules

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Apply KE = 0.5 mv^2 = 249

v^2 = 2* 249/17

v = 5.41 m/s

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Work done   by friction W = u Fd

W   = 0.06 * 17* 9.8 * 6.6 * cos 39

W = 51.27 J

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