Download 2-25. A block of mass m: I.62 kg slides down a frictionless incline

2-25.
A block of mass m: I.62 kg slides down a frictionless incline (Figure 2-A).
The block is released a height h : 3.91m above the bottom of the loop.
(a) What is the force of the inclined track on the
block at the bottom (point A)?
The track in frictionless and hence the law of
conservation of mechanical energy is applicable to
the block. According to this law here if the velocity
of the block at A is vA then
Gain in kinetic energy = loss in potential energy
Or
Or
1 2
mv A  mgh
-------------------- (1)
2
vA  2gh  2*9.8*3.91  8.754 m/s
Now as the block is moving on a circular track it is having a centripetal acceleration and
requires a centripetal force to provide it. If the normal reaction of the track at point A
then the resultant force of the weight of the block mg and the normal reaction N at A will
provide necessary centripetal force and hence we get
mvA2
N  mg 
R
N
Or the force exerted by the track on the block at point A will be
N  mg 

mvA2
v2 
 m g  A 
R
R

vA
Substituting the value of vA from equation 1 we have
mg
2 gh 

 2h 
N  m g 
  mg 1 

R 
R


(b) \t\hat is the force of the track on the block at point B?
Now the radius of the track is R and hence the height of point from the ground is
R - R cos45o =
1 

R 1 

2

vB
NB
Thus the loss in height from the top of the incline will be
h = h Or
1 

R 1 

2

h = h – 0.293 R
Hence by applying the law of conservation of mechanical
energy between top and point B we have
Gain in kinetic energy = loss in potential energy
45o
mg
Or
1 2
mvB  mg h
2
Or
vB  2 g  h  0.293R  m/s
Now again the centripetal force towards the center of the track is provided by the
resultant of normal reaction at B NB and the component of the weight mg of the block in
that direction and hence we have
N B  mg cos 450 
mvB2
R
Solving and substituting value of vB from equation 2 we have
Or
Or
mvB2 mg mg
N B  mg cos 45 


2  h  0.293R 
R
R
2
2h
2h 



N B  mg 0.707 
 0.586   mg  0.121 

R
R



0
(c) At what speed does the block leave the track?
The speed at point B is already calculated in part (b) as
vB  2 g  h  0.293R  m/s
(d) How far away from point A does the block land on level ground?
After leaving the track at point B the block will move under gravity and makes a
projectile motion, projected with speed vB in the direction tangential to the track making
angle 450 with the horizontal at a height R sin450 =
R
from the ground.
2
Applying the formula for the trajectory of a projectile we can calculate the horizontal
distance from point of projection for any height (= - R cos450 in this case) as
y  x tan  
gx 2
2v02 cos 2 
Or
 R * cos 450  x tan 450 
Or

R
Gives 
R
2
2
 x *1 
 x *1 
gx 2
2vB2 cos 2 450
gx 2
2 * 2 g  h  0.293R  *
1
2
x2
2  h  0.293R 
Or
 2  h  0.293R  R  2  h  0.293R  x  x 2
Or
x 2  2  h  0.293R  x  2 R  h  0.293R   0
Solving this quadratic we can calculate the value of x.
Now x is the horizontal distance of landing point from Point B hence the distance from
point A will be
x + R sin 45 = x +
R
2
(e) Sketch the potential energy U(x) of the block. Indicate the total energy on the
sketch.
The potential energy of a body at height h with respect to some reference level is given
by
Potential energy = mgh
As acceleration due to gravity g near the surface of earth for h much less then the radius
of earth is considered constant and equal to 9.8 m/s2, the potential energy is directly
proportional to the height h of the block from the ground (as ground is considered as
reference level where the potential is zero).
Now the height depends on x according to the shape of the track hence on the track the
potential energy graph (verses x) will have the same shape as that of the track and after
point B it will be a projectile hence having the shape of a parabola. So the track will be
given as
U(x)
mgh
0
x
The initial potential energy is mgh and the total energy is also mgh because at that point
the kinetic energy of the block is zero. As the block slides down its kinetic energy
increases and the potential energy decreases according to the graph. At the highest
point of the parabola also it is having some horizontal velocity and hence the potential
energy is less then the total energy. As there is no friction force the total energy remains
constant.