Question 158: Laws of Motion

Question

One end of a mass-less rope, which passes over a mass-less and friction-less pulley $$P$$ is tied to a hook $$C$$ while the other end is free. Maximum tension that the rope can bear is $$360$$ $$N.$$ With what value of maximum safe acceleration (in $$m{s^{ - 2}}$$) can a man of $$60$$ $$kg$$ climb on the rope?

Accepted Answer

Assuming acceleration $$a$$ of the man is downwards.
So the equation will be
$$mg - T = ma$$
$$	herefore$$ $$a = g - {T \over m} = 10 - {{360} \over {60}} = 4$$ $$ m/{s^2}$$
So the maximum acceleration of man is 4 m/s
2
downwards.

Answer

$$16$$

Answer

$$6$$

Answer

$$4$$

Answer

$$8$$

Laws of Motion

One end of a mass-less rope, which passes over a mass-less and friction-less pulley $P is tied to a hook C while the other end is free. Maximum tension that the rope can bear is 360 N. With what value of maximum safe acceleration (in m{s^{ - 2}}) can a man of 60 kg$ climb on the rope?

Choose the correct answer: