Question

If $75\%$ of the energy in part (A) becomes kinetic energy at $\left(B\right)$, calculate the speed at which the skier arrives at $B$.

• A. $15m{s}^{-1}$
• B. $60m{s}^{-1}$
• C. $3m{s}^{-1}$
• D. $30m{s}^{-1}$

Answer 1

Answer: (D)

$30m{s}^{-1}$

Weight of the skier $=$ $60kgf$
Mass of the skier $=$ $60kg$
Initial gravitational P.E. of skier $=$ mgh
$=60\times {10}^{-2}\times 75=$ $45000J$
Final gravitational P.E. of skier $=60\times {10}^{-2}\times 15=9000J$
Decrease in the gravitational P.E. of skier $=(45000-9000)J=$ $36000J$

75% of the P.E. of skier $=\frac{75}{100}36000=27000J$

Now $27000J$ of P.E. is utilised in imparting kinetic energy.
Kinetic energy of the skier $=$$27000J$.
Also kinetic energy of the skier

$=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\times 60\times v^2$

$=\frac{1}{2}\times 60\times v^2=27000J$

$v^2=\frac{27000}{30}$

$=900m^2{s}^{-2}$

v$=\sqrt{900m^2{s}^{-2}}$

$=30m{s}^{-1}$