Question

At a certain factory, 340 kg crates are dropped vertically from a packing machine onto a conveyor belt moving at 1.00 m/s. (A motor maintains the belt's constant speed.) The coefficient of kinetic friction between belt and crate is 0.400. After a short time, slipping between the belt and the crate ceases and the crate then moves along with the belt. Consider the period of time during which the crate is being brought to rest relative to the belt.

Answer 1

Answer:

+b±√b² - 4ac /2a

0.6t ± √36-36/2a

Explanation:

Work done = 1/2 mv² where v =  (1.2)²

Therefore, 1/2m(1.2)ω mgh

1/2m (1.2)² = 0.4 × m ×10 5

s = 1.44 / 2.4 = 1.44 / 8

S = ut - 1/2gt²

Where u = 1.2

g = 0.9 × 10

Therefore,

1.8 = 1.2v-2t²

2t²c-1.2t+1.8 = 0

t² - 0.6t + 0.9 = 0

0.6t ± √36-36/2a

Solving this further, we make use of the formula

+b±√b² - 4ac /2a