Work done during incline - Question

Feb 2020
7
2
Sweden
Hey. I'm wondering if I have answered the following questions correctly.

Question:

Anna first ride a bike (case 1) on a 100 m horizontal distance and then (case 2) she rides up a 50m long slope with a height difference of 7m. Anna weighs 54 kg and her bike weighs 16 kg.

a. How much work is done vertically in case 1? (Neglecting any impact of friction)
b. How much work is done vertically in case 2? (Neglecting friction)
c. What force does the work in question b) ?


Solution:

a. By Newtons second law the vertical resultant force \(\displaystyle F_{y}=0 \iff W_y = F_{y}(100m) =0 \). Since the verticall resultant force is zero no work is performed vertically.

b.

Assume she is travelling with constant velocity. Then the force of gravity which is acting on her and the bike is equal to \(\displaystyle Fg =mg \). Denote the incline angle \(\displaystyle \alpha \). And define the \(\displaystyle {\parallel}\) -axis parallel to the contact surface and the
\(\displaystyle \perp \) axis perpindicular to \(\displaystyle {\parallel}\) -axis. The force of gravity can be broken into its component forces. Now the resultant force \(\displaystyle F_{\perp} =0\) and \(\displaystyle F_{\parallel}=0\) since she is travellling with constant velocity. Denote the force of gravity with is parellel to the contact surface \(\displaystyle Fg_{\parallel}\). By trigonomety it follows that \(\displaystyle Fg_{\parallel}=Sin \alpha \space Fg= Sin \alpha \space mg\). And by newtons second law if the resultant force \(\displaystyle F_{\parallel}=0\) then there exists a force \(\displaystyle Fg^*_{\parallel}\) such that \(\displaystyle Fg_{\parallel}+Fg^{*}_{\parallel}=0\). Then the total amount of work performed while she is travelling up the incline is equal to \(\displaystyle W_{total}=Fg^{*}_{\parallel}(50m)=- Sin \alpha \space mg (50m) \) .

We know that \(\displaystyle g = -9,82 m/s^2\), \(\displaystyle \alpha = Sin^{-1}(7/50) \approx 8,0478 ^{\circ} \) and \(\displaystyle m = 70 kg\). Hence it follows that \(\displaystyle W_{total} \approx 4,8 kJ\).


One can also interpret the question differently. Assume by horizontal they mean the distance she travelled from case 1. Now the force which is vertical to the horizontal line is just \(\displaystyle Fg\). And since the vertical resultant force is zero there must exist an equal and opposite acting force call it \(\displaystyle Fg^{*}=-Fg\). And the work performed in the vertical direction \(\displaystyle W_{vertical} = -Fg(7m)=-mg(7m)\approx 4,8 kJ\). In either case, we obtain the same value.

c.

The forces \(\displaystyle Fg^{*}_{\parallel}\) and \(\displaystyle Fg^{*}\) is the exact amount of force necessary to oppose the equal and opposite directed force of gravity to maintain a constant velocity during the travelling up the incline.
 
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Apr 2015
1,136
294
Somerset, England
Gosh Aion, you are working too hard. Go and take a cold shower.

:)

Yes you are exactly correct

The potential energy gained by the cyclist = the work done (This is called the work energy theorem)


PE = mass x g x height difference = (54+16) x 9.81 x 7 = 4806.9 J

A couple of points

Your question says the "weight..."
This is incorrect and should read the mass of the cyclist and the mass of the cycle.

kg force is a measure of weight but is not an ISO measure so shoudl not be used.

You have not actually said what the force is that does the work as per question, do you want to have another go at this?
 
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Feb 2020
7
2
Sweden
Gosh Aion, you are working too hard. Go and take a cold shower.

:)

Yes you are exactly correct

The potential energy gained by the cyclist = the work done (This is called the work energy theorem)


PE = mass x g x height difference = (54+16) x 9.81 x 7 = 4806.9 J

A couple of points

Your question says the "weight..."
This is incorrect and should read the mass of the cyclist and the mass of the cycle.

kg force is a measure of weight but is not an ISO measure so shoudl not be used.

You have not actually said what the force is that does the work as per question, do you want to have another go at this?
I'm glad im correct. I think my teacher wrote weight by accident. But I'm not sure how to answer question c exactly.
 
Apr 2015
1,136
294
Somerset, England
Work is done when a force moves its point of application along the line of its action.

Is the force of gravity applied at the same point at the beginning and end of the hill climb?
 
Feb 2020
7
2
Sweden
Work is done when a force moves its point of application along the line of its action.

Is the force of gravity applied at the same point at the beginning and end of the hill climb?
I think the force \(\displaystyle \displaystyle Fg^{*}_{\parallel}\) is a constant force which Anna is applying to maintain her constant velocity. And the force \(\displaystyle \displaystyle Fg^{*}\) is constantly applied throughout the line of action but follows from newtons law of gravity and by the second law.
 
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Apr 2015
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I think the force Fg∗∥ is a constant force which Anna is applying to maintain her constant velocity
Since there is no friction why do you think Anna is applying a force to maintain a constant velocity?

What about Newton's first law?

You had this correct in part a.

Once again

Is the force of gravity applied at the same point at the beginning and end of the hill climb?
Remember the question asks for the vertical work

b. How much work is done vertically in case 2?
 
Feb 2020
7
2
Sweden
Since there is no friction why do you think Anna is applying a force to maintain a constant velocity?

What about Newton's first law?

You had this correct in part a.

Once again

Remember the question asks for the vertical work
Am I wrong?

Newtons first law states that the resultant force of all acting forces on an object is zero if the object is either in rest or in motion with constant velocity along a straight line. According to Newton's Second Law, an object affected by different forces will gain acceleration in the same direction as the resultant force. And the resulting force equals mass times the acceleration.


Since the given object has constant speed and direction the acceleration must be zero. And the resultant force on the object must therefore also be zero. When the object is travelling up the incline it is opposing the work which gravity exerts on the object. That work is negative since the force of gravity is acting in the opposite direction of motion. Notice that \(\displaystyle \displaystyle Fg_{\parallel}= Sin \alpha \space mg \) and is acting in the opposite direction of motion. If this was the only force acting on the object in the direction parallel to the contact surface the object would accelerate down the incline by newtons second law. But since we know that the resultant force \(\displaystyle \displaystyle F_{\parallel}=0\) hence there must exist and equally large and opposite directed force \(\displaystyle \displaystyle Fg^*_{\parallel} \) such that their sum is equal to zero.
 
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Apr 2015
1,136
294
Somerset, England
Have you thought about the normal reaction ?

incplane1.jpg
As you say, Newton's laws imply that there is no net force in the vertical direction.

But this also explains why as the incline becomes steeper ( θ increases ) the vertical component of N decreases and it becomes harder to cycle up the plane.

Remember that the net force acts on the body, it is not exerted by the body.

I know that this is a fictional situation as the bicycle will not function without friction, but there you have it.
 
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