Momentum: throwing boot while on ice

[akgtdoskce]

Homework Statement

A 100 kg man sitting on a frozen lake throws a 1.5 kg boot horizontally with a speed of 20 m/s. He is 5m from the shore and the coefficient of kinetic friction is 0.0009. How long does it take for him to reach the shore?

Homework Equations

∑ F = m*a
v2 = v02 + 2a(x − x0)
v = v0 + at
p=mv
F̅ Δt = m Δv

The Attempt at a Solution

The one part I'm really having a problem with is finding the force. I planned on using p=mv to find initial velocity of the man (0.3 m/s), then ∑ F = m*a to find acceleration, v2 = v02 + 2a(x − x0) to find final velocity, and v = v0 + at to find time. However because I'm not clear on how to calculate the force, I'm not able to solve for the other variables. I do believe my teacher mentioned the only force is of friction (mg*µ=.882N) but is it correct to make the assumptions that make this true? Thank you!

 

[gneill]

Homework Statement

A 100 kg man sitting on a frozen lake throws a 1.5 kg boot horizontally with a speed of 20 m/s. He is 5m from the shore and the coefficient of kinetic friction is 0.0009. How long does it take for him to reach the shore?

Homework Equations

∑ F = m*a
v2 = v02 + 2a(x − x0)
v = v0 + at
p=mv
F̅ Δt = m Δv

The Attempt at a Solution

The one part I'm really having a problem with is finding the force. I planned on using p=mv to find initial velocity of the man (0.3 m/s), then ∑ F = m*a to find acceleration, v2 = v02 + 2a(x − x0) to find final velocity, and v = v0 + at to find time. However because I'm not clear on how to calculate the force, I'm not able to solve for the other variables. I do believe my teacher mentioned the only force is of friction (mg*µ=.882N) but is it correct to make the assumptions that make this true? Thank you!

Your teacher is correct. Once the boot is thrown the only force acting on the man (horizontally) is the friction force between him and the ice.

The initial velocity with which he begins his slide is due to the recoil from his throwing the boot. That's where conservation of momentum comes in. No dealing with forces is required there: momentum is conserved no matter what, every time.

 

[haruspex]

v² = v₀² + 2a(x − x₀) to find final velocity, and v = v₀ + at to find time

You can get there a little more directly using $s = v_0 t + \frac 12 at^2$.