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 = 0.04, for a distance d = 8.8 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle = 26° with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of = 26° (thus on the incline it is parallel to the surface) and has a tension T = 31 N.
1) What is the work done by tension before the block gets to the incline?
2) What is the work done by friction as the block slides on the flat horizontal surface?
3) What is the speed of the block right before it begins to travel up the incline?
4) How far up the incline does the block travel before coming to rest?
5) What is the work done by gravity as it comes to rest?
6) During the entire process, the net work done on the block is:

Explanation / Answer

1)

work done by tension = Tension * distance * cos(theta)

work done by tension = 31 * 8.8 * cos(26)

work done by tension = 273.1 J

the work done by tension is 273.1 J

2)

work done by friction = frictional force * distance

work done by friction = 0.04 *(17 * 9.8 - 31 * sin(26)) * 8.8

work done by friction = -53.9 J

3)

using work energy theorum

0.5 * 17 * v^2 = 273.1 - 53.9

v = 5.08 m/s

the speed of the block just at the bottom of the incline is 5.08 m/s

4)

acceleration of the mass on the incline

a = ( - m * g sin(theta) + T)/m

a = (31 - 17 * 9.8 * sin(31))/17

a = -3.22 m/s^2

let the distance is d

using third equation of motion

v^2 - u^2 = 2 * a * d

0 - 5.08^2 = -2 * 3.22 * d

d =4 m

the block will travel 4 m along the incline

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