A block of mass m moves with a velocity of v0 up the plane. The angle of inclination is such that the block is able to slide down. Find the velocity of the block v1, when it returns to the same point at which it was launched. Take mkas the coefficient of friction between the plane and the block.
a. v1
= v0 ((1 - mk ctn q)/(1+
mk ctn q))0.5
b. v1
= v0 ((1+ mk ctn q)/(1
- mk ctn q))0.5
c. v1
= v0 ((1 - mk ctn q)
(1+ mk
ctn q))0.5
d. v1
= v0 ((1 - mk ctn q)
(1+ mk
ctn q))
where ctn is cotangent
The free body diagram
of the block is as shown in the figure.
Consider the forces
parallel to the direction of motion of the block.
When the block is moving
up, the net force in the horizontal direction is
m a0 = m
g sinq + m
g mk cos q
=> a0 =
g (sin q +
mk
cos q)
where a0
is the acceleration of the block when it is moving up.
Consider
the velocity of the block at
A as v0.
The
velocity of block at B is Zero.
v0
= (2 a0
R)0.5
where R is the distance between A and B.
The net force on the
block when it is moving downwards is
m a1 = m g sinq -
m g mk
cos q
=> a1
= g sinq
- g mk
cos q
v1
= (2
a1
R)0.5
where
v1
is the velocity of the block when it returns
to A
v1/v0
= (a1/a0)0.5
v1
= v0
((1 -
mk
ctn q)/(1+
mk
ctn q))0.5
The velocity of the
block when it returns to the same position at which it was launched is
v1
= v0 ((1 - mk ctn q)/(1+
mk ctn q))0.5