Question

A40 kg boy climbs a rope that passes over an ideal pulley. The other end of the rope is attached to a
60kg weight placed on the ground. What is the maximum upward acceleration the boy can have without
lifting the weight? Iſhe climbs the rope with upward acceleration 2 g, with what acceleration the weight
will rise up?​

Answer 1

Answer:

Explanation:

The gravitational force downwards = mg = 40 x 10

= 400N  (taking = 10m/s​²)

Tension T in upwards direction , for maximum acceleration of the boy = T = 60 x10

= 600N

hence T - mg  = ma

600 - 400 = 40 x a

200 /40  = a

a = 5m/s²   in an upward direction.

When the boy is accelerating with acceleration upwards with 2g Let the acceleration of block = a

Equation for boy =  T- mboyg =  mboyx2g ---1

Equation for  block = T- mblock g  =  mblock x a --- 2

Substituting the value of T from eq 1 to 2 -  

​mboy x 2g + ​ mboyg - ​ mblock g  = ​ mblock x a

= 40 x 2 x 10 + 40 x 10 - 60 x10  = 60 x a

= 800 + 400 -600 = 60a

= 1200 -600 = 60a

= 600 = 60a

a = 10 m/s²

Therefore, the acceleration of the weight or block is 10m/s²  in upwards direction.

Answer 2

Answer:

Explanation:

Explanation:

The gravitational force downwards = mg = 40 x 10

= 400N (taking = 10m/s​²)

Tension T in upwards direction , for maximum acceleration of the boy = T = 60 x10

= 600N

hence T - mg = ma

600 - 400 = 40 x a

200 /40 = a

a = 5m/s² in an upward direction.

When the boy is accelerating with acceleration upwards with 2g Let the acceleration of block = a

Equation for boy = T- mboyg = mboyx2g ---1

Equation for block = T- mblock g = mblock x a --- 2

Substituting the value of T from eq 1 to 2 -

​mboy x 2g + ​ mboyg - ​ mblock g = ​ mblock x a

= 40 x 2 x 10 + 40 x 10 - 60 x10 = 60 x a

= 800 + 400 -600 = 60a

= 1200 -600 = 60a

= 600 = 60a

a = 10 m/s²

Therefore, the acceleration of the weight or block is 10m/s² in upwards direction.