An 85 kgman lowers himself to the ground from a height of 10.0 mby holding onto a rope that runs over a frictionless pulley to a 65 kgsandbag. With what speed does the man hit the ground if he started from rest?

1. Mass of a man, $M_m=85\mathrm{kg}$
2. Mass of a sandbag,$M_b=65\mathrm{kg}$
3. Height, $h=10.0m$

Newton’s second law states that the force acting on the object is equal to the product of mass and its acceleration. The direction of the net force is the same as the direction of the acceleration of the object.

Applying Newton’s second law of motion, we can write the equation of motions for a man and the sandbag. Then we can solve those equations to get the desired result.

$V_f^2={\left(V_i\right)}^2+2a\left(y-y_0\right)$ (i)

Here, $V_f$is the final velocity, $V_i$ is the initial velocity, a is the acceleration, yis the final displacement, and $y_0$is the initial displacement.

$F=ma$ (ii)

Here, F is the net force on the object, m is the mass of the object, a is the acceleration of the object.

The motion of the sandbag and man are intheopposite direction, so the equation (ii) can be written as,

$$\begin{gathered} \left({\mathrm{M}}_{\mathrm{m}}\mathrm{g}\right)-\left({\mathrm{M}}_{\mathrm{b}}\mathrm{g}\right)=\left({\mathrm{M}}_{\mathrm{m}}+{\mathrm{M}}_{\mathrm{b}}\right)\mathrm{a} \\ \left({\mathrm{M}}_{\mathrm{m}}-{\mathrm{M}}_{\mathrm{b}}\right)\mathrm{g}=\left({\mathrm{M}}_{\mathrm{m}}+{\mathrm{M}}_{\mathrm{b}}\right)\mathrm{a} \\ \left(85\mathrm{kg}-65\mathrm{kg}\right)9.8\mathrm{m}/{\mathrm{s}}^2=\left(85\mathrm{kg}+65\mathrm{kg}\right)\mathrm{a} \\ \mathrm{a}=\frac{\left(85\mathrm{kg}-65\mathrm{kg}\right)9.8\mathrm{m}/{\mathrm{s}}^2}{\left(85\mathrm{kg}+65\mathrm{kg}\right)} \\ =1.31\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration is $1.31\mathrm{m}/{\mathrm{s}}^2$.

Now, for the velocity after travelling 10 m, we can use kinematic equation (i)

$$\begin{gathered} {\mathrm{V}}_{\mathrm{f}}^2={\left({\mathrm{V}}_{\mathrm{i}}\right)}^2+2\mathrm{a}\left(\mathrm{y}-{\mathrm{y}}_0\right) \\ =\left(0\right)+\left(2\times 1.31\mathrm{m}/{\mathrm{s}}^2\right)\left(10\mathrm{m}/\mathrm{s}\right) \\ {\mathrm{V}}_{\mathrm{f}}=\sqrt{2\times 1.31\mathrm{m}/{\mathrm{s}}^2\left(10\mathrm{m}/\mathrm{s}\right)} \\ =5.1\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the man when he hits the ground is 5.1 m/s.