sputavanomr

2021-11-10

A 60kg skier with an initial speed of 12 m/s coasts up a 2.5 m high rise that makes an angle of $35}^{\circ$ with the horizontal. a) if the coefficient of friction between the skis and the snow is 0.08, find the speed when he reaches the top of the rise. b) how much work does friction do on the skier?

Onlaceing

Beginner2021-11-11Added 15 answers

Part A

First calculate the acceleration

$a=g(\mathrm{sin}0+\mu \mathrm{cos}0)=9.81(\mathrm{sin}35+0.08\mathrm{cos}35)=6.3\frac{m}{{s}^{2}}$

Finally the speed can be found using:

$v=\sqrt{2as+{u}^{2}}=\sqrt{2(-6.3)\left(2.5\right)+{12}^{2}}=10.6\frac{m}{s}$

Part B

First calculate the friction force

$F=\mu mg\mathrm{cos}0=0.08\left(60\right)\left(9.8\right)\mathrm{cos}35=38.5N$

Next multiply the friction by the distance to get the work

W=sF=2.5(38.5)=96.3j

First calculate the acceleration

Finally the speed can be found using:

Part B

First calculate the friction force

Next multiply the friction by the distance to get the work

W=sF=2.5(38.5)=96.3j