Question

The ‘set-up’ shown here, is used to raise a mass M through a height h above the ground. The vertically upward force, F (z) applied by an external agency, varies with the height, z, above the ground as F (z) = F (0) – kz.

The instantaneous speed, acquired by the mass M, when it has been raised through a height h, is given by

(1) \([\frac{2h(F(0)Mg)+kh^2}{M}]^{\frac{1}{2}}\)

(2) \([\frac{2h(F(0))-kh^2}{M}]^{1/2}\)

(3) \([\frac{2h(F(0)-Mg)-kh^2}{M}]^{1/2}\)

(4) \([\frac{2h(F(0))+kh^2}{M}]^{1/2}\)

Answer 1

 (3) \([\frac{2h(F(0)-Mg)-kh^2}{M}]^{1/2}\)

The net upward force, acting on the mass M, when it is at a height z above the ground, is

Let a be the instantaneous acceleration at the height. We have

Ma = F (net)

∴ If v is the instantaneous speed of the mass M when it is at a height h above the ground, we have