Calculating Stopping Time and Distance on a Slope with Kinetic Friction

• quickslant
In summary, 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 the time taken for the skier to come to a stop and the distance traveled down the hill.
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

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

Force = Mass x Acceleration

I have mass but where do i have force?

quickslant said:
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?

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?

quickslant said:
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.

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?

quickslant said:
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.

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

What is kinetic friction?

Kinetic friction is the force that resists the motion of two surfaces in contact when one or both of the surfaces are moving.

What factors affect the amount of kinetic friction?

The amount of kinetic friction depends on the types of materials in contact, the roughness of the surfaces, and the force pressing the surfaces together.

How is kinetic friction different from static friction?

Kinetic friction occurs when two surfaces are already in motion, while static friction occurs when two surfaces are not moving relative to each other. Kinetic friction is typically lower than static friction.

How is kinetic friction measured?

Kinetic friction is typically measured in units of force, such as Newtons or pounds, using a device called a dynamometer. It can also be calculated using the coefficient of kinetic friction, which is a dimensionless value.

How can kinetic friction be reduced?

Kinetic friction can be reduced by using lubricants, such as oil or grease, between the two surfaces in contact. It can also be reduced by using smoother or more slippery materials for the surfaces, or by reducing the force pressing the surfaces together.

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