# Question #ca57c

Jan 11, 2017

Since Gravitational field is a Conservative field, potential energy does not dependent on curved path followed by the crate.

Let $h$ be the vertical height lost by the crate of mass $m$ while moving from Point $A$ to $B$.

Lost of Potential energy $\Delta P E = m g h$
Given $m g = 300 N$, and acceleration due to gravity is $9.81 m {s}^{-} 2$

Change in kinetic energy$\Delta K E = \text{Final kinetic Energy"-"Initial Kinetic Energy}$
$\Delta K E = \frac{1}{2} m {\left(8\right)}^{2} - \frac{1}{2} m {\left(1.2\right)}^{2}$
$\implies \Delta K E = 32 m - 0.72 m$
$\implies \Delta K E = 31.28 m = 31.28 \times \frac{300}{9.81} = 956.57 J$

By Law of Conservation of Energy, this loss gets partly converted into change in kinetic energy and partly is lost as work done against friction ${W}_{f}$. Therefore we have
$\Delta P E = \Delta K E + {W}_{f}$
$\implies {W}_{f} = \Delta P E - \Delta K E$

Inserting calculated values we get
$\implies {W}_{f} = 300 h - 956.57$ $J$

Insert the value of height $h$ (not given in the problem) to calculate the final answer