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

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c) both a and b.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards