solve 3 and 4 using the energy and work 3. A block is given an initial speed of
ID: 2039393 • Letter: S
Question
solve 3 and 4 using the energy and work
3. A block is given an initial speed of 5 m/ s up a 30° incline. The coefficient far up the incline of friction between the block and plane u 0.25. How does the block travel before coming to rest? 4rskg block is pulled along a rough horizontal surface by a force of 20 N directed at a 30° angle above the horizontal. The coefficient of friction between the block and surface is u 0.25. If the block starts from rest, how far along the floor must it be pulled so that the final speed is 4 m/s?Explanation / Answer
Given
3.
initial speed is 5 m/s
the angle of inclinationis 30 degrees
the coefficient of friction is mue = 0.25
from work energy theorem
the work done = change in k.e
0.5*m*(v2^-v1^2) = F_f*s
0.5*m*(v2^-v1^2) = mue*m*g cos theta*s+ mg s sin theta
s = 0.5(v1^2)/(mue*g cos theta+g sin theta)
s = 0.5*5^2 /(0.25*9.8 cos30+9.8 sin 30) m
s = 1.78018 m
the block can travel before coming to rest is 1.78018 m along the incline
4.
given
m = 5 kg , F = 20 N directed 30 degrees above the horizontal
the coefficient of friction is mue = 0.25
the final speed is 4 m/s , initial speed is 0 m/s
the net force force acting on the block is
net force F = Fcos theta - mue(mg - F sin theta)
work done = change in kinetic energy
W = F*S = 0.5*m*v^2
W = (Fcos theta - mue(mg - F sin theta))s
W = (20 cos30 - 0.25(5*9.8 - 20 sin30))s
then change in kinetic energy is k.e = 0.5*5*4^2 J = 40 J
equating these , (20 cos30 - 0.25(5*9.8 - 20 sin30))s = 40 J
solving for s
s = 5.2837 m
the block can move to s = 5.2837 m
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