Question

A 90 kg ice skater moving at 11.0 m/s on the ice encounters a region of rough up ice with a coefficient of kinetic friction 0.500. How far along the rough ice doe she go before stopping?

Answer 1

Kinetic energy=1/2mv2, where m is mass and v is velocity.

Here,initially, m=90 kg and v=11 m/s. So, initially,kinetic energy=1/2*90*11*11=5445 J.

Finally,the object stops. So, kinetic energy=1/2*90*0*0=0 J.

So,change in kinetic energy=final kinetic energy-initial kinetic energy-= 0 - 5445 = -5445 J

Now, frctional force is given by: F=$\mu$N, where $\mu$ is coefficient of friction and N is normal reaction.

In the given poblem,there is no vertical acceleration. So, N=force due to gravity=mg,where m is mass and g is gravitational acceleration.

So, friction F=$\mu$mg=0.5*90*9.8=441 N

Also,work done=fscos$\theta$, where f and s are the magnitudes of force and displacement vectors respectively and $\theta$ is the angle between them.

Now,$\theta$=180 degrees as friction acts opposite to displacement.

So,Work done=441*s*cos180= - 441s

According to work energy theorem, work done=change in kinetic energy

=> - 441s= -5445

=>s=5445/441=12.35 m