A mass m = 13 kg is pulled along a horizontal floor, with a coefficient of kinet

Question

A mass m = 13 kg is pulled along a horizontal floor, with a coefficient of kinetic friction k = 0.05, for a distance d = 7.3 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle = 27° with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of = 27° (thus on the incline it is parallel to the surface) and has a tension T = 25 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

Explanation / Answer

1) Only the tension component in the direction of motion matters:
work = T * cos * d = 25N * cos27º * 7.3m = 162.60 J

2) normal force Fn = mg - Tsin = 13kg * 9.8m/s² - 25N * sin27º = 116.05 N,
so work done by friction is work = -µ * Fn * d = -0.05 * 116.05N * 7.3m = -42.35 J

3) making the net work = 122 J = KE = ½mv² = ½ * 13kg * v²
v = (120.25J / 6.5kg) = 4.30 m/s

4) upslope acceleration = T/m = 25N / 13kg = 1.92 m/s²
downslope acceleration = gsin = 9.8m/s² * sin27º = 4.44 m/s²
making the net acceleration a = 2.52 m/s² downslope
v² = u² + 2as
0 = (4.30m/s)² - 2 * 2.52m/s² * s
s = 3.66 m

5) gravity work = mgdsin = 13kg * 9.8m/s² * -3.66m * sin27º = -125.73J

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