Mechanics net tractive force question

In summary: The friction energy loss is the difference in kinetic energy between the start and end points of the 400m. Your problem refers to climbing 400m up a 1-in-10 hill. That is all the info you need to calculate the rise in height to give you the change in PE. Please show some effort in this thread, or it will be deleted.
  • #1
themster
7
0

Homework Statement


Ascending a 1 in 10 hill, a 1500 kg car slows down from 100 km/h to 50 km/h in a distance of 400 m. Calculate the net tractive force F between the tyres and the road assuming F remains constant and wind resistance can be neglected. Is the driver braking during this period?

I am told the answer is 379N but have no idea how to go about it
any help would be greatly appreciated
 
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  • #2
themster said:

Homework Statement


Ascending a 1 in 10 hill, a 1500 kg car slows down from 100 km/h to 50 km/h in a distance of 400 m. Calculate the net tractive force F between the tyres and the road assuming F remains constant and wind resistance can be neglected. Is the driver braking during this period?

I am told the answer is 379N but have no idea how to go about it
any help would be greatly appreciated

Compare kinetic energy at the start and after the 400m with gain in potential energy. Then look at the energy lost by the friction.
 
  • #3
would i convert the velocities into m/s?
Do I find the differences between the KE and compare that with the PE?

Would you be able to demonstrate it?
 
  • #4
themster said:
would i convert the velocities into m/s?
Do I find the differences between the KE and compare that with the PE?

Would you be able to demonstrate it?

k.e. after = k.e. before - gain in p.e. - friction energy loss.
k.e. = 1/2 m v2 in general
friction energy loss = friction force x distance covered

You are looking for the friction force.

Always use SI units or you'll get the wrong answer:
meter, kilogram, Newton, meters/second, g = 9.81 m/sec2 etc.

Don't know what you mean by "demonstrate it".
 
  • #5
but how to you calculate gain in PE? PE1 - PE2?
I know PE = mgh but what would the value for h be?

I meant worked example by demonstrate it :)
 
  • #6
themster said:
but how to you calculate gain in PE? PE1 - PE2?
I know PE = mgh but what would the value for h be?

I meant worked example by demonstrate it :)

We do NOT do your schoolwork for you here. Re-read the PF Rules if you have any doubt about that (under Site Info at the top of the page).

Your problem refers to climbing 400m up a 1-in-10 hill. That is all the info you need to calculate the rise in height to give you the change in PE. Please show some effort in this thread, or it will be deleted.
 
  • #7
ok sorry i didnot understand it first - i think i was confused by a 1 in 10 hill - what is it?
 
  • #8
themster said:
ok sorry i didnot understand it first - i think i was confused by a 1 in 10 hill - what is it?

1m up for every 10m horizontal
 
  • #9
rude man said:
k.e. after = k.e. before - gain in p.e. - friction energy loss.
k.e. = 1/2 m v2 in general
friction energy loss = friction force x distance covered

You are looking for the friction force.

Always use SI units or you'll get the wrong answer:
meter, kilogram, Newton, meters/second, g = 9.81 m/sec2 etc.

Don't know what you mean by "demonstrate it".


Would the value of h be 400m?
from my working i can't seem to get the answer
I converted 100km/h to 27.8m/s and 50km/h to 13.9m/s
for KE before i get: 1/2 * m * V before^2 = 1/2 * 1500 * (27.8)^2 = 579630
for KE after i get: 1/2 * m * V after^2 = 1/2 * 1500 * (13.9)^2 = 144907.5
for the PE(gain) i get PE=mgh = 1500*9.81*400 = 5886000

using the formula quoted, k.e. after = k.e. before - gain in p.e. - friction energy loss.
i re-arranged to get friction energy loss = -k.e. after +k.e. before - gain in p.e.
from this i get:friction energy loss = -144907.5 + 579630 - 5886000
= -5451277.5

working backwards, using the formula quoted,: friction energy loss = friction force x distance covered, i calculated that the value i need for friction energy loss = 379*400 = 151600.

the only thing i can think is that the height isn't the same as the distance 400m :S
 
  • #10
pe = mgh where h = height. Look at h again. Your last sentence above is a good hint! Surely you know the difference between distance and height ...

Also, the problem asks for the tractive (friction) force, not the friction energy loss.
 

1. What is net tractive force?

Net tractive force is the total amount of force applied to an object in a specific direction, taking into account all the forces acting on the object. In mechanics, it is often used to describe the force required to move an object, such as a vehicle, against resistance.

2. How is net tractive force calculated?

To calculate net tractive force, you must first identify all the forces acting on the object in question. Then, use the formula Fnet = ma, where Fnet is the net tractive force, m is the mass of the object, and a is the acceleration of the object in the direction of motion.

3. What factors affect net tractive force?

The main factors that affect net tractive force include the mass of the object, the force of gravity, the coefficient of friction between the object and its surface, and any additional external forces acting on the object, such as wind resistance or mechanical resistance.

4. Can net tractive force be negative?

Yes, net tractive force can be negative if the forces acting on the object are in opposite directions. This means that the object will experience a net force in the opposite direction, causing it to slow down or move in the opposite direction.

5. How is net tractive force applied in real-world scenarios?

In real-world scenarios, net tractive force is used to understand the motion of objects, such as vehicles or machinery, and to determine the amount of force needed to overcome resistance and achieve a desired velocity. It is also used in engineering and design to ensure that vehicles and machines are able to function properly and efficiently.

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