Answer:

Explanation:
m = Mass of object = 
mg = Weight of object = 20 N
g = Acceleration due to gravity = 
v = Final velocity = 15 m/s
u = Initial velocity = 0
d = Distance moved by the object = 150 m
= Angle of slope = 
f = Force of friction
fd = Work done against friction
The force balance of the system is

The work done against friction is
.
Answer:

Explanation:
Consider two particles are initially at rest.
Therefore,
the kinetic energy of the particles is zero.
That initial K.E. = 0
The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

now, since both the masses have mass m
therefore,

= m/2
The final K.E. of the particles is

Distance between two particles is d and the gravitational potential energy between them is given by

By law of conservation of energy we have

Now plugging the values we get



This the required relation between G,m and d
5 second fall starting at 0 m/s
ball strikes ground at a speed = 49 meters per second.