(Physics) Friction force? Help!?

(a)
By the law of conservation of momentum :
Total momentum just before impact = Total momentum just after impact
5 × 0 + 0.3 × 20 = (5 + 0.3) v
6 = 5.3v
v = 1.1 m/s

Speed of the box and the ball just after impact = 1.1 m/s (to 2 sig. fig.)


(b)
Method 1 :
Initial velocity, u = 1.1 m/s
Final velocity, v = 0 m/s
Displacement, s = 0.70 m

v² = u² + 2as
0 = 1.1² + 2 × a × 0.7
a = -0.86 m/s²

Frictional force, f
= |ma|
= (5 + 0.3) × 0.86
= 4.6 N


Method 2 :
Work done against friction = Decrease in K.E.
f s = (1/2)mu² - (1/2)mv²
f × 0.70 = (1/2) × (5 + 0.3) × 1.1² - 0
Frictional force, f = 4.6 N

Initial momentum = 0.3 * 20 = 6
Total mass = 5.3 kg
5.3 * v = 6
v = 6 ÷ 5.3

This is approximately 1.13 m/s. As the box sides 70 cm, its velocity deceases to 0 m/s. This is caused by the friction. In this problem, the work that is done by the friction force is equal to the kinetic energy of the box and ball.

Work = Ff * 0.7
KE = ½ * 5.3 * (6 ÷ 5.3)^2
This is approximately 3.4 N * m

Ff * 0.7 = ½ * 5.3 * (6 ÷ 5.3)^2
Ff = [½ * 5.3 * (6 ÷ 5.3)^2] ÷ 0.7

This is approximately 4.85 N. I hope this is helpful for you.