Question

a bucket of water of mass 100metric tons accelerates from rest to a stop at 72 km/h on a horizontal track. calculate the work done by the locomotive engine neglecting friction

Answer 1

1. A bucket of water of mass 100 metric tons accelerates from rest to a stop at $72\,\mathrm{km/h}$ on a horizontal track. Calculate the work done by the locomotive engine neglecting friction.

Solution.

$$A = \Delta \frac{m v^{2}}{2} = \frac{m v^{2}}{2} - \frac{m v_{0}^{2}}{2} = \frac{m (v^{2} - v_{0}^{2})}{2}$$

Given: $m = 100\,000\,\mathrm{kg}$, $v = 72\,\frac{\mathrm{km}}{\mathrm{hour}} = 72\,\frac{1000\cdot\mathrm{m}}{3600\,\mathrm{s}} = 20\,\frac{\mathrm{m}}{\mathrm{s}}$, $v_{0} = 0\,\frac{\mathrm{km}}{\mathrm{hour}}$

Find: $A - ?$

$$A = \frac{100\,000\,\mathrm{kg} \cdot \left(20^{2} \left(\frac{\mathrm{m}}{\mathrm{s}}\right)^{2} - 0\right)}{2} = 20\,\mathrm{MJ}$$

Answer:

$$A = 20\,\mathrm{MJ}$$