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Question

The coefficient of kinetic friction is 0.60 (the slope angle is 60°).

What is the acceleration of the block if
(a) it is sliding down the slope, and
m/s2
(b) it has been given an upward shove and is still sliding up the slope?
m/s2

Answer 1

consider the forces on the block:

The force of gravity

The frictional force

When the forces are translated into the frame of reference of the surface of the slope,
the forces parallel to the slope are:

friction:
0.60*cos(60)*m*g
Weight
sin(60)*m*g

note that in order for the block to be sliding down the slope, the parallel component of the weight must be greater than the frictional force
or sin(60) must be greater than .6*cos(60)
which it is

To compute the acceleration,
F=m*a
which is
m*a=sin(60)*m*g-0.60*cos(60)*m*g
divide out the mass
a=g*(sin(60)-0.60*cos(60))
a=5.55 m/s^2

b) For this part, only the force necessary to overcome kinetic friction plus the parallel weight is necessary to keep it moving up the slope. Since it was given a shove and we don't know how much force was applied in the shove, we can only compute the minimum force necessary to over come kinetic friction and weight
F>g*m*(sin(60)+0.60*cos(60))

j

Answer 2

Create a coordinate system, where x is going up the slope and y is perpendicular. We have the weight of the block downwards, which has a horizontal and vertical component
wcos(60) (positive x direction) and -wsin(60) (negative y direction) respectively. Also the normal force N from the slope pushing up on the block. (positive y direction) Also u=kN, where k is the coefficient of kinetic friction and N is the normal force, and u is frictional force. Going down the slope this force is upwards in the positive x direction, resisting the motion of the block. Now, use Newton's second law, which says F=ma. (force equals mass times acceleration). Since we have two directions x and y we will need to break the problem up into x and y directions.So for the forces in the x direction 1. Fx= wcos(60)+u=ma(x), and forces in the y direction 2. Fy=-wsin(60)+N=ma(y). Since there is no acceleration in the y direction,(the block is moving only in the x direction uphill), we can say ma(y)=0 so from 2. N=wsin(60). Now plug that into 2. Since u=kN we get ma(x)=wcos(60) +kwsin(60). So a(x)=wcos(60)+kwsin(60)/m. Since m=w/g,(g is acceleration due to gravity 9.807 m/s2) we get a(x)=g(cos(60)+ksin(60)), now plug in k=.60 and g=9.807m/s2 and you have your answer. If the block is going up the hill then u will be downwards or negative so a(x)=g(cos(60)-ksin(60)). Hope this helps.

Answer 3

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