Question

A boy holds a $40N$ weight at arms length for $10s$. His arm is $1.5m$ above the ground. The work done by the force of the boy on the weight while he is holding it is:

• A. $0J$
• B. $20J$
  
• C. $40J$
  
• D. $15J$
  

Answer 1

Answer: (A)

$0J$

$t=10sec$

Here the work done by a boy on weight is the work done by the tension force from holding hands.

The W.D will be equal to the P.E of the mass if only we are finding the W.D $M$ lifting the mass.

But, to hold the mass at same height , the net work done will be zero.

as, $w=\int \overrightarrow{F}.\overrightarrow{dx}=T\int d\cos \theta$

as $d=0$

$w=0$