Impulse and momentum for an athlete jumper

[Sam Fred]

Homework Statement

Assume that the athlete approaches the take-off line from the left with a horizontal
velocity of 10m/s, remains in contact with the ground for 0.1s, and take off at a 60°angle
with a velocity for 10m/s, determine (1) the vertical component of the average impulsive
force exerted by the ground on his foot. (2) the horizontal frictional force. The mass of
the athlete is 100kg and the coefficient of kinetic friction (μk) is 0.1.

Homework Equations

momentum 1 + ƩF Δt = momentum 2
Friction force = Normal force X μk

The Attempt at a Solution

1)In the vertical Direction
0+(Ry-Weight) Δt = mvsin60
Ry = 9660 N

2)
Friction Force = Ry X μk = 966 N (I think this is wrong)

 

[Doc Al]

The Attempt at a Solution

1)In the vertical Direction
0+(Ry-Weight) Δt = mvsin60
Ry = 9660 N

Good.

2)
Friction Force = Ry X μk = 966 N (I think this is wrong)

Use the same method as you used for part 1. (Is the coefficient of kinetic friction relevant?)

 

[Sam Fred]

So :
m1v1 - Friction Force X Δt = m1v2 cos60
Friction Force = 5000 N ??

 

[Doc Al]

So :
m1v1 - Friction Force X Δt = m1v2 cos60
Friction Force = 5000 N ??

That's right.

 

[Nickie]

I can confirm that your approach and equations are correct. However, the value you calculated for the friction force may be incorrect. The equation for friction force is correct, but the normal force should be the sum of the athlete's weight and the vertical component of the impulsive force exerted by the ground. This would result in a larger normal force and thus a larger friction force. Additionally, it would be helpful to include units in your calculations to ensure accuracy.