College Physics - work energy theorem

Question

A 1500kg self-driving car is traveling with a velocity of 60 m/s when the engine dies and the computer stops working due to power failure. The car travels on 300m of road where the coefficient of kinetic friction is 0.5, before going off road where the coefficient of friction is 1.2. How far does the car travel before coming to a stop as measured from the location where the engine died?

Answer 1

When I go to solve problems such as these, I believe it is good practice to write a list of Givens and what you are going to determine and Find. So:

Given:

m = 1500 kg

v₀ = 60 m / s

D = 300 m

μ_{k, 1} = 0.5

μ_{k, 2} = 1.2

Find:

x = ? m

From engine failer to come to a stop, a.k.a. a speed of zero (v = 0 m / s)

Solution:

So, starting off, we are able to say that all the original energy in this system started as kinetic energy (neglecting a lot of other complicated things that could be going on).

To do the Work-Energy Method and Formula all in one go, you must think of this:

Like already stated, all of the energy starts as kinetic energy (no potential energy because you always at the reference height in this problem/you never change height/altitude) but transforms into the work done by BOTH parts of the road.

So:

Initial: E_i = 1 / 2 m v₀²

Final: E_f = W₁ + W₂

Where work, W, is defined as the following when working with a constant applied force across some displacement (NOT just distance, there's a key difference there) of Δx:

W = F Δx

So:

W₁ = μ_{k, 1}N * D

W₂ = μ_{k, 2}N * d

E_f = N * (D * μ_{k, 1} + d * μ_{k, 2})

Since there was no gain or loss of energy, the initial energy equals the final energy:

1 / 2 mv₀² = N * (D * μ_{k, 1} + d * μ_{k, 2})

The variable being solved for is d, so after a little algebra you'll get:

d = ( mv₀² / 2N - D * μ_{k, 1} ) / μ_{k, 2}

N is just the normal force of the car against the ground and since it is on flat ground, it is the car weight:

N = mg

d = ( v₀² / 2g - D * μ_{k, 1} ) / μ_{k, 2} = ( (60 m / s)² / ( 2 * 9.81 m / s² ) - (300 m) * (0.5) ) / (1.2)

It is good practice to make sure the units cancel out the way they are supposed to so you should double check that and confirm like I did outside of here that they do cancel.

d ≈ 27.9 m

But that is along the portion of the road with the second friction.

x = D + d

x ≈ 327.9 m

I hope this helps.

Answer 2

This is a problem in conservation of energy, recognizing that work is done to overcome friction.

Here, the starting energy is the kinetic energy of the car, which you can find using E = 1/2 mv^2.

Then, the car will do work against friction that is equal to the force of friction (you have the mass of the vehicle and the coefficient of friction) times the distance traveled. So, that is µmgd where m is the mass of the car, g is the acceleration of gravity, and d is the distance traveled. You have µ and d for the road, so you can directly calculate the work done against friction while on the road.

Subtracting that from the starting energy, gives you the energy remaining when the car goes off road:

E_eor = (1/2)m_c(v_c)² - µ₁m_cgd₁

Where

E_eor is the kinetic energy remaining at the point the car leaves the road

m_c is the mass of the car

v_c is the speed of the car when the power fails

d₁ is the distance traveled on the road

µ₁ is the coefficient of friction for travel on the road

Be sure to use correct units.

You can now set the energy (E_eor, above) that is left equal to µ₂m_cgd₂ where µ₂ is for off road friction and d₂ is the distance traveled off road. Vis:

µ₂m_cgd₂ = (1/2)m_c(v_c)² - µ₁m_cgd₁

You can eliminate redundant terms (e.g. m_c), then rearrange and insert actual values to get the value for d₂.

Be sure to add d₁ to d₂ to get d_total as that is what the problem asks for.

Good luck!!