2 MCAT questions - friction

[cbrons]

Homework Statement

1. A car moving at 20 m/s brakes and slides to a stop. If the coefficient of kinetic friction between the pavement and the tires of the car is 0.1, how far does the car slide?

A. 50 m
B. 100 m
C. 200 m
D. 400 m

2. A car moving at 20 m/s brakes and slides to a stop. If the coefficient of kinetic friction between the pavement and the tires of the car is 0.1, how much time is needed for the car to come to a complete stop?

A. 1 s
B. 10 s
C. 20 s
D. 40 s

Homework Equations

U(k)mgcos0

The Attempt at a Solution

Well the answers for both are C, as the back of the book states. The problem is, I don't understand how you can figure out either the time or distance with the information given... can anyone shed some light?

 

[rock.freak667]

Use energy conservation for the first one, the work done by friction is the change in kinetic energy.

 

[kerimek]

I am able to provide an explanation for how to solve these problems using the given information. The key equation to use in these problems is the equation for kinetic friction, which is Fk = μkN, where Fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.

In the first problem, we are given the velocity of the car (20 m/s) and the coefficient of kinetic friction (0.1). We can use the equation Fk = μkN to solve for the force of kinetic friction acting on the car. Since the car is braking and sliding to a stop, the force of kinetic friction is acting in the opposite direction of the car's motion, so we can set up the equation as Fk = -μkN. We also know that the normal force is equal to the weight of the car, which we can calculate using the formula N = mg.

Plugging in the values, we get Fk = -(0.1)(mg). We can then use the equation for acceleration, a = F/m, to solve for the acceleration of the car. In this case, the acceleration is equal to the force of kinetic friction divided by the mass of the car, so we get a = -(0.1)(mg)/m = -0.1g.

Since we know the initial velocity (20 m/s) and the acceleration (-0.1g), we can use the equation for displacement, x = x0 + v0t + (1/2)at^2, to solve for the distance the car slides. In this case, x0 (the initial position) is assumed to be zero, so the equation simplifies to x = (1/2)(-0.1g)t^2. Plugging in the values, we get x = (1/2)(-0.1)(9.8)(t^2) = -4.9t^2.

To find the time it takes for the car to come to a complete stop, we can rearrange the equation to solve for t: t = √(2x/-0.1g). Plugging in the distance of 200 m (since the car slides to a stop), we get t = √(2(200)/-0.1(9.8)) = √(400/-0