Question

27/A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is u 0.5. The distance that the box will move re before coming to rest on it taking g 10 ms2 belt (2) 0.6 m (3) zero (4) 0

answer is 4.pls explain well.

Answer 1

When the box is dropped on the belt initially it will have some acceleration . So after moving some distance it will come to constant velocity as of belt that mean velocity of box will be zero.

Calculation is shown below.

When the box is dropped on the belt initially it will have some acceleration . So after moving some distance it will come to constant velocity as of belt that mean velocity of box will be zero. Calculation is shown below. Figure shows free body diagram of box Forces are mg, downward (due to gravity) N, upward on box by best f, fractional force towards left box moves towards right rightarrow Sum of forces in horizontal direction = 0 f = ma rightarrow And in vertical direction N = mg frictional force, f = mu mg or mu N. mu mg = ma = > a = mu g = 0.5 times 10 = 5 m/s^2 so acceleration of box, a = 5 m/s^2 rightarrow For distance moved before coming to rest At rest means relative velocity of box = 2 m/s so initially initial velocity u = 0, Final velocity V = 2 m/s and acceleration, a = 5 m/s^2 use