M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet
By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
= 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.
The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J
If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN
Answer:
(a) 3750 J
(b) 62.5 kN
Answer:
1. 218.55 N
2. 
3. 
Explanation:
Part 1;
Net force 
where m is mass, g is gravitational force and 
is the angle of inclination
Frictional force, 
is given by
where 
is the coefficient of static friction
Since 
, therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N
Part 2.
Using the relationship that
Frictional force 
The maximum angle of inclination 
Part 3:
Net force on the object is given by
where 
is the coefficient of kinetic friction
= 9.8 ( sin 38 - (0.51) cos 38 )
= 
