Question

A person lifts a 0.1 kg crate hanging over a pulley by walking horizontally, as shown in figure. Asthe person walks 2m the angle of the rope to the horizontal changes from 45° to 30°. How much work (in Joules) does the person do if the crate rises at constant speed.

Answer 1

Answer:

The answer will be  (2-√2)J

Explanation:Let consider a triangle of PBC where A is the mid point of BC.

Now according to the problem the pulley is placed at P pint and the crate of weight 0.1 kg or 1N is hanging from that pulley,

The person lifts that crate by walking.

Now consider the length of PB is l

Now as the persons walking the angle of the rope with the horizontal changes.

Therefore  for △PAB  tan45 = l/AB => AB = l

Similarly for △PCB tan30 = l/BC => BC = √3l

According to the problem the man walks 2m

Therefore  BC -AB = 2

                => √3l -l = 2

                 => 2l^2 = 2

                 => l = 1

Therefore AP = √AB^2 +PB^2 = √1^2+1^2 = √2

                  CP = √BC^2+ PB^2 = √(√3)^2 +1^2 = 2

Therefore the work done = (2-√2) x 1 x cos 0

                                            =  (2-√2)J

Answer 2

Answer:

bhai sahi solution mile toh mujhe bhi batana khoj raha hu kabse :(