You are helping your friend move, using a ramp to move boxes from the ground to

Question

You are helping your friend move, using a ramp to move boxes from the ground to the moving truck. You give a 24kg box a shove so it moves 1.5m/s at the bottom of the ramp. The angle that the ramp makes with the ground is 12(degrees). The coefficients of static and sliding friction are 0.35 and 0.1, respectively.

a)For how much time does the box slide up the ramp?

b)How far up the ramp does the box slide?

c)When the box stops sliding up the ramp, does it remain stopped or begin sliding back down?

d)Explain how you determined your answer for part c.

Explanation / Answer

Part A)

Net force = ma

umgcos(angle) + mgsin(angle) = ma (mass cancels)

(.1)(9.8)(cos 12) + (9.8)(sin 12) = a

a = -2.996 m/s2 (Negative since it will decelerate)

Then apply vf = vo + at

0 = 1.5 + (-2.996)(t)

t = .501 sec

Part B)

d = vot + .5at2

d = (1.5)(.501) + (.5)(-2.996)(.501)2

d = .376 m up the ramp (37.6 cm)

Part C)

The force down the plane will be mgsin(angle) = 24(9.8)(sin 12) = 48.9 N

The frictional force is umg(cos angle) = (.35)(24)(9.8)(cos 12) = 80.5 N

Since the frictional force is greater than the force down the plane, the box will remain stopped

Part D)

Explained above...

Since the frictional force is greater than the force down the plane, the box will remain stopped

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