To avoid an accident a driver steps on the brakes to stop a 1000-kg car travelling at 65 km/h. The braking distance is 35 m, how much force is necessary to stop the car?

Answer 1

In this case the frictional force stops the car and it could be found with the use of Newton's second law

f = ma,

here m - mass of the car and a - its acceleration ( better to say deceleration, because the speed of the car is 0 in the end).

Acceleration we can find, if apply the formula from kinematics

a = v₀² / 2d,

where v₀ = 18 m/s - initial speed of the car ( I converted km/h to m/s),

d = 35 m - stopping distance.

Now we see that force

f = mv₀² / 2d.

Plugin in the values of all variable, we get

f = 10³ kg· (18 m/s)² / 2·35 m

f = 4.6 x 10 ³ N = 4.6 kN

Answer: 4.6 kN

Answer 2

Let's first solve for a_x in terms of μ and g.

We know that ∑F_x = ma_x = -F_k (Sum of Forces in the x-direction equals m*a equals negative force of Kinetic Friction (know from FBD)).

We also know that ∑F_y = ma_y = 0 = N_y - mg. Therefore N_y = mg.

F_k = μ * N_y = μ * mg

so ma_x = -(μ * mg) or a_x = -(μ * g)

To solve for the force of Kinetic Friction, we first need to solve for μ.

Let's use the Kinematic Equation (V_f)² = (V₀)² + 2ad

If we substitute a for -(μ * g), and plug in the respective values for V_f, V₀, and d.

[V_f = 0] [V₀ = 65 km/hr or 18.0556 m/s] [d = 35m]

And solve we get the value for μ as .47474.

Force of Kinetic Friction is (μ * Normal Force) or (μ * mg) (we know this from N2L with respect to ∑F_x and ∑F_y).

So to solve for the Force of Kinetic Friction which is the only force acting on the car, we can plug in the values and end up with F_k = (μ * mg) = 4657.21 N.

FBD of the car in case anyone needs it:

^ N

F_k <--- CAR

∨ W = mg