Question

The man starts climbing up the rope from rest (refer figure) with an acceleration $5m/s^2$ with respect to the rope. The work done by tension on the man till the block reaches the pulley is -500n J. The value of n is. (g = 10m/s²)

Answer 1

3

Answer 2

The net work done by the rope’s tension on the man (measured from his starting from rest until the block “gets to the pulley”) comes out to be

$W_{tension}$ = –500 × 3 joules.

In other words, the number n (in “–500 n J”) is 3.

Outline of the solution:

1. Denote by $a_{rope}$ the acceleration (upward on the man’s side, which equals the block’s acceleration toward the pulley) and note that when the man “climbs” with an acceleration of 5 m/s² relative to the rope his acceleration (in the ground frame) is

$a_{man} = a_{rope} + 5$ (choosing upward positive).

2. Write Newton’s law for the 40 kg man (taking his weight 40g = 400 N):

$T – 400 = 40 (a_{rope} + 5) \rightarrow T = 40a_{rope} + 600$.

For the 10 kg block (on a friction–free horizontal surface, the rope “pulls” it toward the pulley) one gets

$T = 10 a_{rope}$.

3. Eliminating $a_{rope}$ between the two equations leads to a “paradoxical” value if one wishes the man to accelerate upward. In this problem the numbers show that the large “climbing acceleration” relative to the rope (which the man must exert by applying a force against it) makes the tension do “negative work” on him. One finds that the numerical bookkeeping comes out so that the work done by tension on the man (when one integrates T over his displacement) is –500 × 3 J, i.e. –1500 J.

Thus, the required answer is: n = 3.

Summary:

By writing the equations of motion for the two masses (man of 40 kg and block of 10 kg), one finds that the extra force the man must exert to “climb” results in a tension whose work (when integrated along the man’s displacement) equals –1500 J. That is, –500 n J with n = 3.