Question

A bullet of mass 0.0016 kg moving at 499 m/s impacts a large fixed block of wood and travels 6.7 cm before coming to rest. Assuming that the deceleration of the bullet is constant, find the force exerted by the wood on the bullet.

Answer in units of kN

A 4.1 kg bucket of water is raised from a well by a rope. The acceleration of gravity is 9.81 m/s^2 .
If the upward acceleration of the bucket is 2.1 m/s2 , find the force exerted by the rope on the bucket of water.
Answer in units of N

Answer 1

1.

V_f² = V_i² + 2ad
0 = (499 m/s^2 )^2 + 2 a (.067 m)
a = -1.858 E6 m/s^2 [negative just says it's decreasing in acceleration]

E_f = ma
E_f = .0016 kg x (-1.858E6) m/s^2
E_f = -2.9728E3 kg m /s^ == -2.9728E3 N
That is the force the bullet exerts on the block. However the force of the bullet on the block is the same force as the block on the bullet.

= 2.9728 kN [ I reversed sign to help distinguish the direction of the force, the bullet is negative while the block on the bullet is positive]

2.

Newton's second law

{Since the bucket is accelerating upwards, the resultant force is in the upward direction, which means Tension(force exerted by the rope on the bucket) > Weight
Hence,
Tension(T) - Weight = ma (in accordance to Newton's Second Law of Motion)
T - mg = ma
T = ma + mg }

sum of forces = m a

Tension in rope - weight = m a

T = m g + ma = m(g+a)

T = 4.1 kg (9.8m/s/s + 2.1m/s/s) = 48.79N

= 48.8 N