Physics-a couple energy problems...

[Xe54]

1.) In the 1950s, an experimental train that had a mass of 2.50x10^4kg was powered across a level track by a jet engine that produced a thrust of 5.00x10^5 N for a distance of 509m.

A-Find the work done on the train.
B-Find the change in Kinetic Energy.
C-Find the final kinetic energy of the train if it started from rest.
D-Find the final speed of the train if there were no friction.
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A...W=Fd....W=(5.00e^5N)(509m)....W=2.5e^8.

B,C, and D I cannot understand. For part b, I believe you use KE=1/2mv^2, but I have no velocity..., if I had a quantity for velocity, I would be able to find ΔKE by doing K1=...,K2=.... and finally...W=K2-K1=.....J.

For c and d, I'm totally lost.
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2.) A 14,700N car is traveling at 25m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force between the tires and the road is 7100N. How far will the car slide once the brakes are applied?
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Well, I know this:
F=14,700 N
Vi=25 m/s
Vf=0 m/s
avg. braking force=7100 N
and I'm assuming that I need to find d=......?

What formulae would I use to find the answer? I know these:

KE=1/2mv^2
W=ΔKE
W=fd. I'd be able to solve for d, but how would I go about finding the value for W?

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[loveoforganic]

For part b: use F/m = a, then go to kinematics, then 1/2mv^2

For part c: should be the same as b?

For part d: kinematics

 

[MintJulep]

For the second question, does it state anything about a friction force being present?

 

[IntelInside]

1.) In the 1950s, an experimental train that had a mass of 2.50x10^4kg was powered across a level track by a jet engine that produced a thrust of 5.00x10^5 N for a distance of 509m.

A-Find the work done on the train.
B-Find the change in Kinetic Energy.
C-Find the final kinetic energy of the train if it started from rest.
D-Find the final speed of the train if there were no friction.
-------
A...W=Fd....W=(5.00e^5N)(509m)....W=2.5e^8.

B,C, and D I cannot understand. For part b, I believe you use KE=1/2mv^2, but I have no velocity..., if I had a quantity for velocity, I would be able to find ΔKE by doing K1=...,K2=.... and finally...W=K2-K1=.....J.

For c and d, I'm totally lost.
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2.) A 14,700N car is traveling at 25m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force between the tires and the road is 7100N. How far will the car slide once the brakes are applied?
----------
Well, I know this:
F=14,700 N
Vi=25 m/s
Vf=0 m/s
avg. braking force=7100 N
and I'm assuming that I need to find d=......?

What formulae would I use to find the answer? I know these:

KE=1/2mv^2
W=ΔKE
W=fd. I'd be able to solve for d, but how would I go about finding the value for W?

since loveoforganic answered #1, I will answer #2

Well if we look at the diagram for #2 (after you draw it) we will see the kinetic friction force (the braking force) is pointing in the opposite direction to our relative velocity (lets call the direction of velocity the positive direction). Now lets get our variables and values we know listed:

Vi = 25m/s
Vf= 0m/s
acceleration due to friction = -7100/1500 = -4.75 m/s^2 (approximately)
t=?
x=?

Now we have to find out our displacement from the point of where we first apply the brakes.

What linear motion formula would you use? First off we dont know time and x. This we dont even have to worry about if we use

Vf^2=Vi^2+2*a*delta(x)

0=25^2+2*-4.75*delta(x)
-625 = -9.5*delta(x)

Delta x = approximately 65 m

 

[KD1655]

The net work done on/by a system is equal to the change in kinetic energy. Therefore, your answer to b is going to be the same as your answer to a, assuming that there is no friction on the system. I believe that you can make that assumption on the MCAT unless they explicitly refute that in the question. To find the final kinetic energy:

W= 1/2m(vfinal)^2-1/2m(vinitial)^2

Since you as starting from rest, vinitial is zero so the equation reduces to:

W=1/2m(vfinal)^2

Therefore, the question to c will be the same as the asnwers for a and b.

Use the above equation to find (vfinal). In this problem, speed and velocity are interchangeable since the displacement and the distance are equivalent.

Okay for the second problem:

The weight of the car is 14,700 N. Remember that weight is a force and is equal to m*g. From that relationship, you can find the mass of the car to be something like 1470 kg. In this problem, the normal force is equal and opposite to the weight of the vehicle (-14700 N). Since we know what the mass, initial and vinal velocity is, we can calculate the work using the relationship that work is equal to the change in kinetic energy. At that point, you can use the fact that the force that is stopping the vehicle (braking force) is -7100 N (Since it is being applied in the direction opposite of the motion of the car), the work calculated by the change in kinetic energy by the W=F*d equation.

Doing so:

KE(final)= 0 since v=0
KE initial= 1/2(1470)(25)^2 = 459,375 J

W= 0-459,375 = -459,375 J

W=-459,375 J = F *d= (-7100 kg)d

d= 64.7 m

Hope this helps, if you have any further question, PM me for some more specialized assistance.

 

[IntelInside]

Okay for the second problem:

The weight of the car is 14,700 N. Remember that weight is a force and is equal to m*g. From that relationship, you can find the mass of the car to be something like 1470 kg. In this problem, the normal force is equal and opposite to the weight of the vehicle (-14700 N). Since we know what the mass, initial and vinal velocity is, we can calculate the work using the relationship that work is equal to the change in kinetic energy. At that point, you can use the fact that the force that is stopping the vehicle (braking force) is -7100 N (Since it is being applied in the direction opposite of the motion of the car), the work calculated by the change in kinetic energy by the W=F*d equation.

Doing so:

KE(final)= 0 since v=0
KE initial= 1/2(1470)(25)^2 = 459,375 J

W= 0-459,375 = -459,375 J

W=-459,375 J = F *d= (-7100 kg)d

d= 64.7 m

Hope this helps, if you have any further question, PM me for some more specialized assistance.

This brings up a good point that these problems can be solved multiple ways. One using linear motion equations (as I did) and using work and energy as KD1655 did

 

[Geekchick921]

since loveoforganic answered #1, I will answer #2

Well if we look at the diagram for #2 (after you draw it) we will see the kinetic friction force (the braking force) is pointing in the opposite direction to our relative velocity (lets call the direction of velocity the positive direction). Now lets get our variables and values we know listed:

Vi = 25m/s
Vf= 0m/s
acceleration due to friction = -7100/1500 = -4.75 m/s^2 (approximately)
t=?
x=?

Now we have to find out our displacement from the point of where we first apply the brakes.

What linear motion formula would you use? First off we dont know time and x. This we dont even have to worry about if we use

Vf^2=Vi^2+2*a*delta(x)

0=25^2+2*-4.75*delta(x)
-625 = -9.5*delta(x)

Delta x = approximately 65 m

This is how I solved this one as well.