Question

A truck of mass $500kg$ moves from the rest at the top of a section of track $400m$ long and $30m$ high, as shown. The frictional force acting on the truck is $250N$ throughout its journey.
What is the final speed of the truck?

• A. $14m{s}^{-1}$
• B. $24m{s}^{-1}$
• C. $31m{s}^{-1}$
• D. $190m{s}^{-1}$

Answer 1

The correct option is A $14m{s}^{-1}$

Consider the problem

From the conservation of energy,

GPE of truck at top $=mgh=500\times 9.81\times 30=147150J$

Work done against friction $F\times d=250\times 400=100000J$

$GPEoftruckattop=KEoftruckatbottom+workagainstfriction$

$147150=KEoftruckatbottom+100000$

$\begin{array}{c}KEoftruckatbottom=147150-100000=47150J\\ \frac{1}{2}m{v}^2=47150\end{array}$

Final speed $V=\sqrt{\frac{2\times 47150}{500}}=13.7m{s}^{-1}\approx 14m{s}^{-1}$

Hence, Option $A$ is the correct answer.