Calculating Stopping Time and Distance on a Slope with Kinetic Friction

[quickslant]

A sight on many "bunny" hills across ontario is young skiers pushing on ski poles and gliding down a slope until they come to rest. Observing from a distance, you note a person 25kg pushing off with the poles to give an initial velocity of 3.5 m/s. If the inclination of the hill is 5 degrees and coeffecient of kinetic friction of skis on dry snow is 0.20, calculate

a)time taken for skier to come to a stop

b)the distance traveled down the hill

i need help with the A part and I am not sure where id start

 

[radou]

Write down Newton's 2nd law. You can calculate the acceleration from it, and then answer the question by means of simple kinematics.

 

[quickslant]

Force = Mass x Acceleration

I have mass but where do i have force?

 

[radou]

Force = Mass x Acceleration

I have mass but where do i have force?

Have in mind that you are writing Newton's 2nd law for the direction of motion, i.e. the direction of the incline. So, what are the forces acting on the skier in the direction of the incline?

 

[quickslant]

ok.. tell me if this is possible
If i want to find the force that is acting on the skier i can use
Fk = (mu)k Fn
where Fn is mg cos5
when i find out the force can i then calculate acceleration and go from there?

 

[radou]

ok.. tell me if this is possible
If i want to find the force that is acting on the skier i can use
Fk = (mu)k Fn
where Fn is mg cos5
when i find out the force can i then calculate acceleration and go from there?

That was the force of kinetic friction. You have one more force acting on the skier. After finding that force, find the resultant force (i.e. the sum of these two forces). Then you can use Newton's 2nd law to find the acceleration.

 

[quickslant]

the other force is gravity's components which is mg + mg sin5?

if i add those two together and subtract frictional force will i have a resultant force? and will i be able to go from there?

 

[radou]

the other force is gravity's components which is mg + mg sin5?

if i add those two together and subtract frictional force will i have a resultant force? and will i be able to go from there?

What do you mean by 'mg + mg sin(5)'? The 'other' force we were talking about is the component of gravity along the incline, and it equals mg sin(5). Now, as you said, subtract the force of friction from this force and you'll have a resultant force.

 

[quickslant]

i included mg as force of gravity but i forgot its canceled out by the force normal..
thanks