Question

A block of mass m is thrown vertically down form a height of 8.0 meters. Determine the mass of the block, its initial and final velocities, and the potential and kinetic energies at the start and end of its travel. The kinetic energies of the block when it is 2m below the start is 250J and the 642 J when it is 2m above the ground.

Answer 1

Explanations & Calculations

• Refer to the sketch attached

• By the kinetic energies given it could be written,

$\qquad\qquad \begin{aligned} \small E_2&=\small 250J\\ \small 250&=\small \frac{1}{2}.m.V_2^2\\ \small V_2^2&=\small \frac{500}{m}\cdots(1)\\\\ \small E_3&=\small 642J\\ \small 642&=\small \frac{1}{2}.m.V_3^2\\ \small V_3^2&=\small \frac{1284}{m}\cdots(2) \end{aligned}$

• Considering the motion within this period along with the motion equation $\small V^2=U^2+2as$,

$\qquad\qquad \begin{aligned} \small V^2_3&=\small V^2_2+2g\times4\\ \small V_3^2-V_2^2&=\small 8g\\ \small \frac{1284}{m}-\frac{500}{m}&=\small 8g\\ \small m&=\small \frac{784}{8\times9.8}\\ &=\small \bold{10\,kg} \end{aligned}$

• By applying the same equation for the first 2m & last 2m, those velocities could be calculated.

$\qquad\qquad \begin{aligned} \small V_2^2&=\small V_1^2+2gs\\ \small V_1&=\small \sqrt{\frac{500}{10kg}-2\times9.8\times2}\\ &=\small \bold{3.29\,ms^{-1}}\\\\ \small V_4^2&=\small V_3^2+2gs\\ \small V_4&=\small \sqrt{\frac{1284}{10kg}+2\times9.8\times2}\\ &=\small \bold{12.95\,ms^{-1}} \end{aligned}$