In a ballistics test, a 28-g bullet pierces a sand bag that is 30cm thick. If the initial bullet velocity was 55 m/s and it emerged from the sandbag moving at 18m/s, what was the magnitude of the friction force (assuming it to be constant and the only force present) that the bullet experienced while it traveled through the bag

Explanation:

Given that,

The mass of the bullet is

M= 28g = 0.028kg.

Thickness oof sack bag

d = 30cm = 0.3m

Initial velocity of bullet Vi = 55m/s

Velocity at which the bullet emerges the sack bag is Vf = 18m/s

We want to calculate the frictional force present in the sack bag.

Using conservation of energy.

Work done by friction in the sack is equal to change in kinetic energy.

Work done by friction is given as

Wf = Fr × d

The distance is the thickness of the bag

Wf = Fr × 0.3

Wf = 0.3 • Fr

Change in kinetic energy is calculated by

∆K.E = ½mVf² — ½ mVi²

∆K.E = ½m(Vf² — Vi²)

∆K.E = ½ × 0.028 ( 18²-55²)

∆K.E = 0.014 × -2701

∆K.E = -37.814 J

Since

Work done by friction = ∆K.E

0.3 Fr = -37.814

Fr = -37.814/0.3

Fr = -126.05 N

The, negative sign show that frictional force is opposing the motion.

Fr = 126.05 N

v=ir = 3x15 =45 v