What is the skis co-effecient of kinetic friction with the ice?

1. Feb 5, 2013

Bostonpancake0

Sam, whose mass is 75kg takes off down a 50m high, 10° slope on his jet powered skis. The skis have a thrust of 200N Sam's speed at the bottom of the slope is 40m/s. What is the skis co-effecient of kinetic friction with the ice?

2. Relevant equations

Thought originally to use conservation of energy at top and bottom, then calculate energy loss, then find distance of slope using trigonometry thus allowing me to use w=fxd to find the frictional force acting on skis, then use Fk=μxmass, to co-effecient of kinetic friction, but soon noticed he gained energy down the slope due to his thrust! stuck??

2. Feb 5, 2013

Simon Bridge

The thrust does work on Sam doesn't it?
So you could use conservation of energy or kinematics for this.

3. Feb 5, 2013

tms

Draw a diagram. Show the forces on the skier. Then write down the kinematics equations.

4. Feb 5, 2013

Bostonpancake0

Yes, the thurst does work, and I'm still not getting the correct answer, could someone please show me working??

5. Feb 5, 2013

tms

Show your work so people can see where you are going wrong.

6. Feb 5, 2013

Bostonpancake0

This is my working

File size:
34.2 KB
Views:
300
7. Feb 5, 2013

Simon Bridge

in equation (1) - how do you work out the friction force?

Last edited: Feb 5, 2013
8. Feb 5, 2013

haruspex

You've multiplied the coefficient of friction by Sam's weight. Is the normal force equal to Sam's weight?

9. Feb 5, 2013

Bostonpancake0

sorry yes thats my bad equation 1 should have (co-effecient of kinetic friction)xsams normal force which is not his weight 75 but rather 735xcos(10°) which now gives a co-effecient of 0.165 which is correct thank you i would never have seen that haha

10. Feb 5, 2013

Simon Bridge

No worries - we can all get blind to odd little things sometimes.