Mechanics question please help

Question

A car whose engine can work at a maximum rate of 20 kW is traveling along a straight horizontal road. The maximum speed of the car is 40 m/s. Find the resistance to motion.

The car, which is of mass 750 kg, comes to a hill of inclination sin^-1 (0.1) . Assuming the resistance to be proportional to the speed, find the maximum steady speed, at which the car can climb the hill. Find also the retardation when the car is moving up the hill at a speed of 25 m/s.

Answer 1

Hi Orawan!

 

To obtain an answer for the problem in the first paragraph, I assume:

 

a.  The engine is exerting its maximum power

b.  The car is moving at its maximum speed, which is reasonable to associate with its maximum power

c.  By "find the resistance to motion," it is meant that we should calculate the net force resisting motion

 

We can go into the whole Newton's third law description of the car pushing on the road and the road pushing back on the car, but -- for ease of description -- we can paraphrase that situation as the power of the engine provides a force that pushes the car forward

 

In this case, we can use the average power formula, that:

 

P = Fv

 

where 

 

P = applied power

F = applied force

v = speed

 

(this is readily derived from the expression for work (W) that:  W=Fd, where d is the magnitude of displacement the force causes;  this expression is valid as long as the force is exerted directly along the line of the displacement.  Thus, (ΔW/Δt) = Δ(Fd)/Δt = F(Δd/Δt) if the force is a constant "average" force.  (ΔW/Δt) = P and (Δd/Δt) = v, thus P = Fv)

 

So, if the car is moving at its maximum speed of 40 m/s, and the engine is providing the power, we can calculate the force pushing the car forward, using P = Fv.

 

Then, if we realize the car is moving at a constant 40 m/s, we can use Newton's second law to show that, in the direction of motion:

 

F_net = F (applied) + F_resist = ma = 0  (since the speed is constant, a = 0)

So now we have a direct relationship between the applied force and resisting forces.

 

[NOTE:  the NET work done on the car as it moves is zero, because the car does not change speed, and thus does not change kinetic energy -- and the work-energy theorem says that net work must equal change in kinetic energy.  So the resisting forces do negative work -- and thus take away energy -- at the same rate the engine provides it]

 

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In the second part, with the car on the incline, we can make the following suppositions, based on what we are told and have calculated:

 

a.  To achieve its maximum possible constant speed, the engine must be providing its maximum power, and thus applying the same force calculated in the previous section

b.  The resisting force is directly proportional to speed, meaning F_resist = kv, where k is a constant

c.  By "retardation" in the last sentence, they mean the total forces acting against the car driving up the hill when v = 25 m/s.

 

We have a way of calculating k, because we calculated the value of the resisting force when v= 40 m/s.  So solving for k becomes a one-step algebra problem.

 

Then, we can set up Newton's 2nd law along the direction of the slope.  In this direction, the forces acting are the applied force of the engine (assumed maximum, for the reasons above), the resisting force (which depends on whatever speed the car is traveling on the slope), and the component of gravity pulling the car back down the plane (which has formula F_dp = mgsin(θ), where θ is the incline of the slope to the horizontal).

 

We are told θ = sin^-1(0.1), meaning sin(θ) = sin(sin^-1(0.1)) = 0.1.

 

So Newton's 2nd law in the direction of the slope becomes (with down the slope taken to be negative):

 

F_net = F (applied) - F_resist - mgsin(θ) = F (applied) - kv - (750kg)(9.8m/s²)(0.1) = ma = 0 (since speed is steady, so a = 0)

 

Since F (applied) has been solved for in the previous part, and k can be solved for as mentioned above, this becomes one equation with one unknown, v -- the maximum steady speed the car can attain up the incline.

 

The total retardation forces holding the car back as it drives up the hill at the moment v = 25 m/s can then be calculated by computing F_resist for that speed and adding it to F_dp, which is independent of speed.

 

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If you have any further questions about this problem, or would like to check an answer, just let me know.  I hope this discussion gets you on track!