In the situation described in the problem a considerable amount of kinetic energy of the bullet emerging out of the gun will be used up for the bullet to get embedded in the block of wood. In all probability, in a real situation, most of the energy of the bullet will be used up this way. However, the the question provides no data on the energy used up this way. Therefore to solve the problem we assume that the entire kinetic energy of the bullet has been used up in moving the block of wood horizontally against force of friction.
Further it is assumed that the block starts moving with the speed of muzzle speed of bullet just as it is hit by it and then decelerates at uniform rate till it comes to a stop.
Given:
Mass of bullet = m1 = 15 g = 0,015 kg
Mass of wood block = m2 = 1.24 kg
Coefficient of friction = M = 0.28
Distance moved by block = d = 11.0 m
We know acceleration due to gravity = g = 9.81 m/s^2
Calculating muzzle speed of bullet:
The normal force (fn) exerted by the block on horizontal surface is given by:
fn = m2*g
And frictional force of block (f) against the horizontal surface is given by:
f = fn*M = m2*g*M
Work done or energy expended (e) in moving the block is given by:
e = f*d = m2*g*M*d = 1.24*9.81*0.28*11 = 37.466352 J
This energy is supplied by the kinetic energy of bullet which is given by the formula
Kinetic energy of bullet = e = 1/2*m1*v^2 = 1/2*0.015*v^2 = 0.0075v^2
Where v = muzzle speed of the bullet
Therefore:
0.0075v^2 = 37.466352
Or: v = (37.466352/0.0075)^1/2= 70.6789 m/s
Calculating declaration of block:
deceleration = a = (v^2)/(2*d) = 70.6789^2/(2*11) = 227.0688 m/s^2
Answer:
Muzzle speed of bullet = 70.6789 m/s
Deceleration of block = 227.0688 m/s^2
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