Question

work done to stop 5kg mass moving at 30 m/s

Answer 1

Answer:

Since the pulleys are assumed to be massless, so forces on the movable pulley balance-

2T=T

′

.

Also,T∗a is constant

So a

′

comes out to be

2

a

.(a is acceleration of 5kg block,and a

′

is acceleration of 10kg block)

Equation of motion for the 5kg block and 10kg block are as follows-

T−5gsin30

o

=5a

10g−2T=10

2

a

From these equations,a comes out to be

3

g

ms

−2

a=3.3msec

−2

Answer 2

Answer:

 a 5kg mass slips over a plane inclined 30o with horizontal. What will be its acceleration when it moves upwards. Other mass is 10kg. Assume that there is no friction anywhere. (g=10m/sec2)