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