0
$\begingroup$

Question:

A man of weight W hangs at the end of a light extensible rope whose modulus is $nW$. The other end of the rope is fastened to a fixed point. He proceeds to climb up the rope. Prove that when he reaches the fixed point, he has done $\frac{2n+1}{2n+2}$ times the work he would have done in climbing the same distance up an extensible rope.

We can easily calculate the total length of string to be $l+\frac{l}{n}$ where $l$ is natural length of string.

But in my book it is written that the work done by the man in climbing up the rope = work done against gravity - Work done by the tension in the string = Wl (l+n)/n - 1/2 .w . l/ n I am completely helpless to understand that why work done against grevatity is wl (l+n)/n . I think it should be Tl (l+n)/n where T is force applied by person to climbing up . I think $T > W$ because $T = W$, then it should be position of equilibrium. When $T=W$, the person should not move. Please help me. This concept is totally new to me.

$\endgroup$
0
$\begingroup$

In explaining how the work done against gravity is indeed $W*(l+l/n)$. Think of it this way; what would be the work done against gravity if you had to lift a block of weight $W$ through $h$ meters? It would have to be its increase in Gravitational Potential Energy(GPE) which is $Wh$ as any more force greater than $W$ would cause the block to not be stationary. Hence it’s the same for the man on the rope. His work done against gravity is his increase in GPE( Note how GPE uses a force which is very close or approximately equal to the weight that can just lift the object through a distance).

Note:You mistakenly put an extra $l$ in the energy equation giving it units of $Nm^2$ but it instead should be just be $W*(l+l/n)$ so it has units of $Nm$ as energy is = force x distance moved in direction of force

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.