A bike is traveling on a horizontal surface with friction and is making a sharp

Question

A bike is traveling on a horizontal surface with friction and is making a sharp turn. When it is making a turn of radius 11.3 meters, it is going the fastest it can travel to maintain that radius.   It then increases his speed and skids out to radius of 33.1 meters (he is also traveling the fastest he can go to maintain this new radius of 33.1 meters here as well).   If the mass of the bike is 57.9 kg and coefficient of static friction is 0.56, what is the net work done by all the forces acting on the bike in Newtons as it skidded from the radius of 11.3 meters to 33.1 meters?

Explanation / Answer

Here ,

for intial velocity is u

as

centripetal force = static frictional force

m * u^2 /11.3 = 0.56 * m * 9.8

u = 7.87 m/s

Now , let the final speed is v

centripetal force = static frictional force

m * v^2 /33.1 = 0.56 * 9.8 *m

v = 13.5 m/s

Now ,

using work energy theorum

net work done = change in kinetic energy

net work done = 0.5 * 57.9 *(13.5^2 - 7.87^2)

net work done = 3600 J

the net work done by all forces on the bike is 3600 J

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