If a force of 10 N acts on an object and an additional force of 6 N acts on the object in the same direction. What will be the net force be on the object?
4N
60 N
6N
16 N

D

Explanation:

For this kind of problem, forces add. F = F1 + F2

F1 = 6 N

F2 = 10 N

F = 6N + 10N

F = 16N

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

hin = L = 0.7 m

hfin = 0 m   ⇒    Ufin = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*hin = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*hin) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)