A box of mass m = 1.5 kg is dropped vertically onto a conveyor belt which is moving with a constant speed v = 6.2 m/s. The coefficient of kinetic friction between the box and the conveyor belt is: µk = 0.12. How much time does it take before the box attains the same speed as the conveyor belt? t =

t = 5.27s

Explanation:

F (net) = m.a(1)

friction force = μ . N

friction force = μ . m.g(2)

equate eqn 1 and eqn 2

a =  μ . g

a = 0.12 × 9.8

a = 1.177m/s²

kinematic equation

V = v + at

t = 6.2 / 1.177

t = 5.27s

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge