ONONDAGA. Physics 103 Exam 2 1. A stockroom worker pushes a box with mass 10.9 k

Question

ONONDAGA. Physics 103 Exam 2 1. A stockroom worker pushes a box with mass 10.9 kg on a horizontal surface with a constant speed of 2.60 m/s. The coefficients of kinetic and static friction between the box and the surface are 0.270 and 0.480, respectively. What horizontal force must the worker apply to maintain the motion of the box? If the worker stops pushing, what will be the acceleration of the box? If the worker starts again pushing, after the box came to rest, what minimum horizontal force must he exert in order to start it moving? 2. One straightforward way to measure the coefficients of friction between a box and a wooden surface is illustrated in the figure. The sheet of wood can be raised by pivoting it about one edge. If the angle reaches the value 550, the box just begins to slide downward. (a) How large is the coefficient of static friction between the box and the surface? (b) After the box starts sliding, its acceleration down the incline is a 1.5 m/s If the mass of the box is 1.2 kg, how large is the friction force? (c) Find the value of the coefficient of kinetic friction between the box and the surface. Plywood sheet 3. Consider the system shown in the figure. Block A has weight 4.91 N and block B has weight 2.94 N. Once block B is set into downward motion, it descends at a constant speed. Assume that the mass and friction of the pulley are negligible. (a) Calculate the coefficient of

Explanation / Answer

1) a) Force to keep moving with constant speed = ukmg = 0.27 x 10.9 x 9.8 = 28.8414 N

b) Accelaration =  -ukg = -0.27 x 9.8 = -2.646 m/s2

c) minimum force required =  usmg = 0.48 x 10.9 x 9.8 = 51.2736 N

2) a) mgsin(theta) =  usmgcos(theta) =>  us = tan55 = 1.428

b) mgsin(theta) - friction = ma => friction = 1.2(9.8sin55 - 1.5) = 7.833 N

c) friction =  ukmg =>  uk = 7.833/(1.2 x 9.8) = 0.666

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