Let us consider block of mass $m$ hanging by rope with force $T$. When no other horizontal and vertical forces are acting on a block, the block is at equilibrium if $$T=mg.$$ And if the rope is pulled up a distance $h$, $$\text{work done}=mgh$$

Don't we need more force to move the block upward? If we just use $T$, the body will be in equilibrium. Don't we need more force to lift the block?

  • 2
    $\begingroup$ Possible duplicate of How can we move an object with zero velocity? $\endgroup$ – Aaron Stevens Feb 23 '19 at 4:55
  • $\begingroup$ @AaronStevens Should I rescind my vote if I think this question is a duplicate? $\endgroup$ – PiKindOfGuy Feb 23 '19 at 5:01
  • $\begingroup$ @PiKindOfGuy It's up to you $\endgroup$ – Aaron Stevens Feb 23 '19 at 5:01
  • $\begingroup$ @AaronStevens Okay, thanks. I just wasn't sure if there was standard protocol. $\endgroup$ – PiKindOfGuy Feb 23 '19 at 5:02
  • $\begingroup$ Yes okay the question is same but not kindda duplicate $\endgroup$ – Roshan Feb 23 '19 at 6:13

Yes and no. We need to pull upwards for a tiny tiny tiny amount of time with a force just a tiny tiny tiny bit greater than $T$ and then if we maintain a force $T$, the block will be in equilibrium and will be traveling at a constant velocity upwards.

  • $\begingroup$ That means we neglect the small force that is making the body move? $\endgroup$ – Roshan Feb 23 '19 at 6:15
  • $\begingroup$ In calculating the work? Yes because we can make the work done by that force as small as we'd like. $\endgroup$ – PiKindOfGuy Feb 23 '19 at 6:45

Consider Newton's law on inertia. He states that a body tends to remain in its state of rest or motion unless there was a net force on the body.

Here the block has no net force acting on it since the tension $T$ cancels out the weight $mg$ of the block, hence it remains in a state of rest, but if we apply an infinitesimally small force upwards we can produce a net force acting upwards for a small time which causes it to move upwards.

Now Newton's Law takes over again and body moves up even without any external force (assuming $T$ to be constant).


You do need the force exerted by the string to be greater than $mg$ at the end the force exerted by the string needs to be less than $mg$.

Assume that the block is the system.

At the start let the external force applied by the string to the block be $mg +t$.

This force is applied for a distance $d$.

The work done by this external force on the block is $(mg+t)d$.

Now change the external force to $mg$ and move the block up a distance $h-2d$ so the work done by the external force is $mg(h-2d)$.

Towards the end apply a force of $mg-t$ for a distance $d$ which means that the work done by the external force is $(mg-t)d$.

So the total work done by the external forces is (mg+t)d+ mg(h-2d) + (mg-t)d =mgh$.

What is happening is that at the start the net force on the block is the sum of the force due to the string $mg+t$ (upwards) and the force due to the gravitational attraction of the earth $mg$ (downwards) so the net force acting on the block is $mg+t+(-mg) =t.

In moving a distance $d$ the total work done on the block is $mgd$ and this results in the block gaining kinetic energy (work energy theorem).

For the next part of the motion the net force on the block is $mg+(-mg)=0$ and so the kinetic energy of the block does not change.

Towards the end the motion the net force acting on the block is $mg - t +(-mg) = -t$ and so the work done on the block is $-td$ ie there is a loss in kinetic energy.
This loss in kinetic energy towards the end is the same magnitude as the gain in kinetic energy at the start so the net gain in kinetic energy is zero ie if the block started at rest it finished up at rest.

In summary all that need to be done is to make sure that $t_{\rm start}d_{\rm start} - t_{\rm end}d_{\rm end} =0$ and the way this is often explained is to have an infinitesimal increase in the tension at the start and a corresponding infinitesimal decrease in tension at the end.

  • $\begingroup$ Okay you made this understandable . But if we consider a body is never going to stop (say light). For that body also we neglect small force causing it to move.. don't we. If not, will the workdone be whole new expression? $\endgroup$ – Roshan Feb 23 '19 at 11:53
  • $\begingroup$ @Roshan I have assumed that you were asking about matter not light for which this analysis is wholly inappropriate. $\endgroup$ – Farcher Feb 23 '19 at 12:27
  • $\begingroup$ Yes i was asking for matter just gave wrong example $\endgroup$ – Roshan Feb 23 '19 at 12:30
  • $\begingroup$ @Roshan I your question the block started from rest. If the Bloch was already moving upwards then an upward force of $mg$ acting on the block will allow it to move with constant velocity. $\endgroup$ – Farcher Feb 23 '19 at 12:38

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