1 a.)A box is sliding down an incline tilted at a 14° angle above horizontal. Th
ID: 1597932 • Letter: 1
Question
1 a.)A box is sliding down an incline tilted at a 14° angle above horizontal. The box is initially sliding down the incline at a speed of 1.5 m/s. The coefficient of kinetic friction between the box and the incline is 0.37. How far does the box slide down the incline before coming to rest? (Give answer to the nearest 0.01 meters).
b.)
A 4.6-kg block rests on a = 25° incline as shown in the figure. If the coefficient of static friction between the block and the incline is 0.71, what magnitude horizontal force F in Newtons must act on the block to start it moving up the
incline? (Give answer to the nearest 0.1 N).
c.)
An object weighing 3.7 N falls from rest subject to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b = 3.7 N s2/m2. What terminal speed will this object approach? (Give answer to the nearest 0.01 m/s).
Explanation / Answer
Given
1
a)
Box is sliding down the incline of 14 degrees above the horizontal
initial velocity is u =1.5 m/s
coefficient of kinetic friction is mue =0.37
writing the force acting on the box along horizontal is
mg sin theta - mue_k*mg cos theta = ma
a = g( sin theta -mue_k*cos theta)
a = 9.8(sin14 - 0.37 cos14) m/s2
a = -1.14746 m/s2
-ve sign indicates the accelerataion is decreasing means the box is slowing down and finally stops after soem displacement
using equations of motion v^2 -u^2 = 2as
s = -u^2/2a
s = -1.5^2/(2*(-1.14746)) m
s = 0.9804263 m
the box slide down the incline before coming to rest is 0.9804263 m
b) mass of block m= 4.6 kg, theta = 25 degrees
coefficient of static friction between the block and the incline is mue_s = 0.71
the forces acting on the block along horizontal in order to move up is
the force required is F = mg sin theta+mue_S *mg*cos theta
F = mg( sin theta +mue_s cos theta)
F = 4.6*9.8(sin25 +0.71 cos25)
F = 48.06 N
c)
weight of the object w = 3.7 N
drag force Fd = bv^2
when the object reaches the terminal speed the weight = drag force
W = Fd
W = bv^2
v^2 = W/b
v = sqrt(W/b)
v = sqrt(3.7/3.7) m/s
v = 1 m/s
the terminal speed is v = 1m/s
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