A car moves up a hill of incline 1 in 50 with a velocity of 15kmph. The resistance due to friction is $\frac{1}{25}$th of weight of the car. The speed
of the car while moving down hill using same power is

If we visualize the truck moving uphill as depicted in the illustration, the restoring force acts downward in opposition to the weight acting downward.

$\sin \theta =\frac{1}{50}$

$P=F_1{v}_1=F_2{v}_2=same \_\_\_\_\_\_\left(1\right)$

for upward motion,

$$\begin{gathered} F_1=\frac{Mg}{25}+mg\sin \theta \_\_\_\_\_\left(2\right) \\ =\frac{Mg}{25}+\frac{Mg}{50}=\frac{3}{50}Mg \end{gathered}$$

for downward motion,

$F_2=\frac{Mg}{25}-\frac{Mg}{50}=\frac{Mg}{50}\_\_\_\_\_\left(3\right)$

$\frac{3}{50}Mg{v}_1=\frac{Mg}{50}{v}_2$

$v_2=3v_1$

Hence, we have the truck's downward speed as =

$3\times 15=45km/hr.$