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Mechanical energy problem help needed!

  • Thread starter hendrix7
  • Start date
  • #1
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Homework Statement


A car, of mass 1100 kg, is travelling down a hill, inclined at an angle of 5o to the horizontal. The driver brakes hard and skids 15 m. The coefficient of friction between the tyres and the road is 0.7. Find the initial kinetic energy and speed of the car.


Homework Equations


work done = force x displacement
weight of car down the hill = 1100 x 9.8 x sin 5o
normal reaction of road with car = 1100 x 9.8 x cos 5o
friction force = 0.7 x 1100 x 9.8 x cos 5o

The Attempt at a Solution


I really do not know how to go about solving this. This braking hard and skidding has stumped me.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
472
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Hi there,

The skidding just implies that a constant breaking.

Cheers
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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Hi hendrix7! :smile:
Yes, a car can either roll or skid.

If it rolls, then no energy is lost (the point of contact, of the wheel with the ground, is stationary, so the friction does no work).

Only if it skids is energy lost. :wink:
work done = force x displacement
weight of car down the hill = 1100 x 9.8 x sin 5o
normal reaction of road with car = 1100 x 9.8 x cos 5o
friction force = 0.7 x 1100 x 9.8 x cos 5o
That's right … just carry on :smile:
 
  • #4
13
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tiny-tim

sorry if I appear dumb, but I still don't understand. Does it matter if the car skids forwards or sideways?
 
  • #5
tiny-tim
Science Advisor
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Does it matter if the car skids forwards or sideways?
oh yes, it does matter …

but I think we can assume that the question means its only skids forwards

(basically, a skidding car is just a block of rubber sliding down the hill! :wink:)
 
  • #6
472
0
Hi there,

In the idea, I would say it does not matter which way the car slide. As long as the coefficient is considered to be the same.

Cheers
 
  • #7
13
0
Okay, here is my attempt at a solution. I would welcome any input from you guys as to whether my method is sound.
Firstly, I have to assume that the car was travelling at constant speed before the braking so that I can then calculate the tractive force of the engine:
weight of car down the hill + tractive force = friction with road
Reaction with road = 1100 x 9.8 cos 5o = 10,739 N
Weight of car down the hill = 1100 x 9.8 x sin 5o = 939.5 N
939.5 + tractive force = 0.7 x 10,739 = 7,517.3
so the tractive force of the engine before braking = 7,517.3 - 939.5 = 6577.8 N
When the car begins to skid, there is no friction between the car and the road. The only forces acting on the car are the braking friction and the weight of the car. Therefore, the work done by the brakes against the engine must be 6577.8 x 15 = 98,667 j and this must be the k.e. of the car before the brakes were applied.
1/2 x 1100 x v2 = 98,667
v = 13.4 m s-1
 
  • #8
472
0
Hi there,

Why don't you start from the start and use Newton's second law correctly:[tex]\sum \vec{F} = m\vec{a}[/tex]

From there, which forces are acting on your vehicle. From these forces, you can determine the acceleration ([tex]\pm[/tex]) of the car.

Assuming a constant acceleration, you can apply the simple equations of a uniformly accelerated motion. And there you go, you have your solutions.

Hint: don't forget that you have two dimensional vectors here.
 
  • #9
Sorry to revive a dead thread, but I'm currently looking at this problem and am just wondering why the weight of the car down the hill is

1100 x 9.8 x sin 5o = 939.5 N
 
  • #10
olivermsun
Science Advisor
1,244
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Sorry to revive a dead thread, but I'm currently looking at this problem and am just wondering why the weight of the car down the hill is

1100 x 9.8 x sin 5o = 939.5 N
First analyze the units. The car is 1100 kg. Acceleration due to gravity is 9.8 m/s^2. Force is in Newtons, which is kg m/s^2 (check this using F = ma). So the units are correct.

The sin 5° comes about because gravity acts straight down. You can split this F into two components -- downhill and into (normal to) the hill -- which are just F sin 5° and F cos 5°. You can remind yourself which is which by thinking of a "flat" hill, where all the force due to gravity is "into" the hill and none is "downhill." Since sin 0° = 0 and cos 0° = 1, the sin must be downhill and the cos must be into the hill.

Hence the original problem, which is the "weight" (gravity force) of the car down the hill:

1100 kg * 9.8 m/s^2 * sin 5°.
 

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