Will a Passenger Slide Off the Seat Without a Seat Belt?

[Sean77771]

Homework Statement

A car traveling at 20 m/s stops in a distance of 50m. Assume that the acceleration is constant. The coefficients of friction between a passenger and the seat are u_s = 0.5 and u_k = 0.3. Will a 70 kg passenger slide off the seat if not wearing a seat belt? Prove it.

Homework Equations

f_s(max) = u_s*N
x_f = x_i + v_i*t + 1/2 at^2

The Attempt at a Solution

I used the first equations to find that f_s(max) is equal to 343N and the second to find that the acceleration is -4m/s^2. I don't know where to go from here. Please, I need this for class tomorrow morning! Thanks.

 

[Dick]

The force acting on the passenger to decelerate him is the static friction of the seat. How much acceleration can a force of up to 343N deliver to a 70kg passenger? As much as 4m/sec? (Put units on numbers, please, makes things clearer).

 

[Sean77771]

The force acting on the passenger to decelerate him is the static friction of the seat. How much acceleration can a force of up to 343N deliver to a 70kg passenger? As much as 4m/sec? (Put units on numbers, please, makes things clearer).

That's the problem, I don't know how to find that out.

 

[Dick]

That's the problem, I don't know how to find that out.

Uh, F=ma?

 

[Sean77771]

Haha, should've known that one. Ok, so a comes out to be 4.9m/s^2, which is greater than 4m/s^2. So does that mean that the seat can hold the passenger up to 4.9m/s^2, therefore he does not slide off the seat? Where does u_k come in?

 

[Dick]

Just as you say, static friction is enough to hold the passenger. If you were to push the passenger so he starts sliding then you would use u_k to conclude kinetic friction isn't enough to stop him from continuing to slide.

 

[Sean77771]

Ok, great, thanks so much for your help!