A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 80 N at an angle of 25 degree above the horizontal. The box has a mass of 25 kg, and the coefficient of kinetic friction between box and floor is 0.300. a) Find the acceleration of the box b) The student now starts moving the box up a 10 degree incline, keeping her 80 N force directed at 25 degree above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?(adsbygoogle = window.adsbygoogle || []).push({});

Thanks for the patience for reading the problem, my work so far is:

a) For the y component,

Fy = ma, since a is 0...

Ty+Ny-Wy = 0

80sin25+Ny-25*9.8 = 0

Ny = 211N

fx(friction) = uNy = 0.300*211N = 63.4

For the x component,

Fx = ma

Tx-fx = ma

80cos25-fx = 25a

72.5-63.4 = 25a

a = 0.364m/s2. This is my answer, but I'm not sure whether it's right or not, but the problem begins in part b.

b) For the y component

Fy = ma

Ny + Ty + Wy = 0

Ny + 80sin25-245cos10 = 0

Ny = 207.5

f = uN = 0.300 * 207.5 = 62.2N

For the x component.

F = ma

Tx-Wx-fx = ma

80cos25 - 245sin10-62.2 = 25a

a = -1.29m/s2

I do not believe the acceleration to be negative in the incline. If so, the student cannot pull the box up and the problem cannot have taken place. I checked my work for like 10 times. Can someone check what I did wrong? Thanks

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# Homework Help: A force-related problem

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