Read the paragraph and answer the questions given below
In physics, a force is said to do work if, when acting on a body, there is a displacement of the point of application in the direction of the force. For example, when a ball is held above the ground and then dropped, the work done on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement). potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
(1)A body of mass 0.3 kg is taken up an inclined plane of length 10 m and height 5 m and then allowed to slide down to the bottom again.The coeficient of friction between body and the plane is 0.15
what is work done by gravitational force over round trip?
what is work done by applied force on upward journey
what is work done by frictional force over the round trip
what is work done by kinetic energy at the end of the trip
(2)A body of mass 1 kg is allowed to fall freely under gravity.Find the momentum and the kinetic energy of the body after 5 sec.after it starts falling
(3)A body of mass 40 kg is dropped from an aeroplane at a height 1 km above the ground.what is its KE (i)at end of 10 sec.(ii)on reaching the ground
(4)If g is the accelearation due to gravity,gain in PE of an object of mass m raised from surface of the earth to a height equal to radius R of the earth is
(5)A bullet of mass a and velocity b is fired into a large block of mass c.the final velocity of the system is?

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its a big paragraph but only 3 questions are there
please answer them


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1)  work done = 0 by gravitational force, as displacement is 0
    work by applied force on upward journey = 0.3 kg * g * 5 m
   length of inclined plane:  slope = 5√5 m
   cos θ = 10/5√5 = 2/√5
   work by frictional force:   μ m g cos θ * (2 *5√5)
               = 0.15 * 0.3 * g * 2/√5 * (10√5)   Joules         =  9 J
   work is not done by kinetic energy... work is done by force.
   KE = m g h - work by friction during sliding down of the body
         = 0.3 * 10 * 5 -  4.5  J
     v = u + g t = 5*10 m/sec
   momentum = 1 kg * 50 = 50 kg-m/s
   v = u + g t = 10 * 10 sec = 100 m/s
       KE at 10 sec = 1/2 * 40 kg * 100²  Joules
       KE at ground level = m g h = 40 kg * 10 * 1000 m = 0.4 MJ
If the height above Earth's surface is small, then change in PE = m g h

if the height h  is significant, then
     change in PE on Earth = G M m h / (R* R1) = m g h * R / (R+h)
                                           = m g R/2    if h = R

2 5 2
sir for 5th one answer is different
ans is ab/a+c
please solve it again
thankyou sir...................
apply conservation of linear momentum