Work done by a force down a ramp

In summary, the conversation discusses how to calculate the average accelerating force from point X to Y, taking into account the mass of the bike and person, their velocity at point Y, and the distance traveled along the ramp. There is also a question about the role of gravity in changing GPE and how the accelerating force affects it. It is noted that the problem may require additional information, such as the time taken to travel from X to Y, in order to fully solve it.
  • #1
38
2
Homework Statement
Calculate the average accelerating force down the ramp.
Relevant Equations
W = fx
E = mgh
1683305127274.png

The mass of the bike and person is 190kg
Calculate the average accelerating force from X to Y, if the bike has a velocity of 30 at point Y.
I am struggling with this question, I know that Fx = Work Done, but I also know that the only way to release GPE as KE is for gravity to do positive work on the object (for the object to fall), so in this case the work done by gravity is 190 x 9.8 x 25, so the change in GPE is -(190 x 9.8 x 25), and the KE is 1/2 x 190 x 900, but I am unsure how the accelerating force can cause a change in GPE, as I always thought that it was gravity that does that. Could someone help my understanding please?
 
Physics news on Phys.org
  • #2
I think the "accelerating force" is responsible for adding the kinetic energy above what would be gained in considering just gravitational work from X to Y.
 
  • #3
nav888 said:
... but I am unsure how the accelerating force can cause a change in GPE, as I always thought that it was gravity that does that.
What makes you think that the accelerating force changes the GPE? What you always thought is correct. Only the gravitational force changes the gravitational potential energy. The accelerating force changes the kinetic energy of the motorcycle.

What is the complete statement of the problem? Does the bike start from rest at the top of the ramp?
 
  • #4
And can I add...

nav888 said:
...if the bike has a velocity of 30 at point Y.
'30' what units? mph? km/h? m/s? ...

nav888 said:
I am struggling with this question, I know that Fx = Work Done
Worth noting that this requires 'x' to be measured in the direction of the force.

nav888 said:
but I also know that the only way to release GPE as KE is for gravity to do positive work on the object (for the object to fall), so in this case the work done by gravity is 190 x 9.8 x 25, so the change in GPE is -(190 x 9.8 x 25), and the KE is 1/2 x 190 x 900
Since the values are different, this tells us that one or more other forces (apart from gravity) is/are acting.

nav888 said:
, but I am unsure how the accelerating force can cause a change in GPE, as I always thought that it was gravity that does that. Could someone help my understanding please?
If we had, for example, a falling stone with no air resistance, then gravity (the only force) would be the accelerating force and the reduction in GPE would equal the gain in KE. But this is not the situation here.

If the motocyclist accelerates rapidly from X to Y the time taken is short.
If the motocyclist accelerates slowly from X to Y the time taken is long.
But the change in GPE is same in both cases.

And, as hinted at by @kuruman, are you sure the question is complete? I would have thought that we also need the time taken to get from X to Y.
 
  • #5
Steve4Physics said:
And, as hinted at by @kuruman, are you sure the question is complete? I would have thought that we also need the time taken to get from X to Y.
Google image search finds the following:
https://www.thestudentroom.co.uk/showthread.php?t=5403130 said:
attachment.png


i) Calculate loss of GPE from X to Y = [elided]

ii) Calculate Kinetic energy at Y = [elided]

iii) The total distance travelled by the motorbike from X to Y along the ramp is 120 m. Calculate the average accelerating force that the motorbike engine provides along the ramp.

@haruspex -- do you want to point out that the problem setter is glossing over the difference between an average over time and an average over distance?
 
Last edited:
  • Like
Likes Steve4Physics and erobz
  • #6
jbriggs444 said:
@haruspex -- do you want to point out that the problem setter is glossing over the difference between an average over time and an average over distance?
And an ill-defined distance at that.
 
  • Like
Likes erobz
  • #7
jbriggs444 said:
Google image search finds the following:@haruspex -- do you want to point out that the problem setter is glossing over the difference between an average over time and an average over distance?
Sometimes I wonder if the posters leave out pertinent information just to torment us.

I assumed it was the average over the distance, but it wasn’t clear how crude of an average they were intending.
 
  • Like
Likes MatinSAR
  • #8
kuruman said:
And an ill-defined distance at that.
Do you mean approximately a quarter of a circle with a radius of 25 meters isn’t 120 meters? 😉
 
  • #9
erobz said:
Do you mean approximately a quarter of a circle with a radius of 25 meters isn’t 120 meters? 😉
I had not seen the 120 m distance when I posted #6.

On edit:
I would be more comfortable with this problem if it asked "Calculate the force that the motorbike engine provides along the ramp assuming that it is constant."
 
Last edited:
  • Like
Likes jbriggs444
  • #10
kuruman said:
I had not seen the 120 m distance when I posted #6.

On edit:
I would be more comfortable with this problem if it asked "Calculate the force that the motorbike engine provides along the ramp assuming that it is constant."
Yeah, @jbriggs444 just found it. I was just making a joke. But I agree.
 
  • #11
jbriggs444 said:
@haruspex -- do you want to point out that the problem setter is glossing over the difference between an average over time and an average over distance?
You knew that I would not resist.
The usual flaw that grabs attention
Rears its head in one dimension
The second adds a twist.

"Average force" is defined as a time average, ##\frac{\int\vec F.dt}{\int dt}##, but let's accept we are here asked for an average over distance. Since the force (static friction) is necessarily in the direction of travel, ##dW=\vec F.\vec{ds}=F.ds##. But force is a vector, so the average force (over distance) is ##\frac{\int\vec F.ds}{\int ds}##.

The force has to supply the energy gap ##E=\frac 12mv^2-mgh##. If the trajectory is straight down distance ##h## with ##F=0 ## then straight across distance ##x## with ##F=F_x## then ##E=F_xx## and the average force is ##\frac{(F_xx, 0)}{x+h}=(\frac E{s},0)##. (##s## being the total distance ##x+h##.)
At the other extreme, the force is only applied during descent. ##F=F_y##, ##E=F_yh## and the average force is ##\frac{(0, F_yh)}{x+h}=(0, \frac E{s})##.

What if we further modify the question to "magnitude of average (over distance) force"?
Exercise for the reader: prove that this is ##\frac Es## regardless of the trajectory and how the force varies over it.
But note how this differs from "average magnitude of .. etc.". Show that this does depend on such details.
 
  • Like
Likes jbriggs444
  • #12
haruspex said:
You knew that I would not resist.
May the average force be with you.
 

Suggested for: Work done by a force down a ramp

Replies
4
Views
975
Replies
59
Views
2K
Replies
4
Views
1K
Replies
4
Views
189
Replies
4
Views
482
Replies
6
Views
1K
Replies
9
Views
793
Replies
28
Views
277
Back
Top