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An object of mass 1kg slides 3m down a roof inclined at 20 degrees to the horizontal. The object then falls 5m under gravity to hit the ground with speed 8m/s. Find the frictional force between the tile and the roof.
The answer given in the book is 9.42N.
Here's where I'm stuck, assuming no resistance forces act while falling, the kinetic energy gained by the object is 50 Joules. However the kinetic energy at 8m/s is 32 Joules. I can't find the speed at which the object just starts to fall.
Thanks.
The answer given in the book is 9.42N.
Here's where I'm stuck, assuming no resistance forces act while falling, the kinetic energy gained by the object is 50 Joules. However the kinetic energy at 8m/s is 32 Joules. I can't find the speed at which the object just starts to fall.
Thanks.
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#2
(Original post by Ethan7702)
An object of mass 1kg slides 3m down a roof inclined at 20 degrees to the horizontal. The object then falls 5m under gravity to hit the ground with speed 8m/s. Find the frictional force between the tile and the roof.
The answer given in the book is 9.42N.
Here's where I'm stuck, assuming no resistance forces act while falling, the kinetic energy gained by the object is 50 Joules. However the kinetic energy at 8m/s is 32 Joules. I can't find the speed at which the object just starts to fall.
Thanks.
An object of mass 1kg slides 3m down a roof inclined at 20 degrees to the horizontal. The object then falls 5m under gravity to hit the ground with speed 8m/s. Find the frictional force between the tile and the roof.
The answer given in the book is 9.42N.
Here's where I'm stuck, assuming no resistance forces act while falling, the kinetic energy gained by the object is 50 Joules. However the kinetic energy at 8m/s is 32 Joules. I can't find the speed at which the object just starts to fall.
Thanks.
Trying to do it via considering the speeds is going to be rather involved at best, since you don't know what the final vertical velocity is. It's probably doable, but I wouldn't fancy trying it via that method.
Since we're talking energy, I'd use conservation of energy. Works out easily - I assume you're using g=10 m/s/s.
Edit: Applied to the whole motion from start to hitting the ground.
Last edited by ghostwalker; 1 year ago
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(Original post by ghostwalker)
No idea where you got the 50 Joules from.
Trying to do it via considering the speeds is going to be rather involved at best, since you don't know what the final vertical velocity is. It's probably doable, but I wouldn't fancy trying it via that method.
Since we're talking energy, I'd use conservation of energy. Works out easily - I assume you're using g=10 m/s/s.
Edit: Applied to the whole motion from start to hitting the ground.
No idea where you got the 50 Joules from.
Trying to do it via considering the speeds is going to be rather involved at best, since you don't know what the final vertical velocity is. It's probably doable, but I wouldn't fancy trying it via that method.
Since we're talking energy, I'd use conservation of energy. Works out easily - I assume you're using g=10 m/s/s.
Edit: Applied to the whole motion from start to hitting the ground.
I still can't wrap my head around the process, could you please give me some pointers?
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#4
(Original post by Ethan7702)
I thought the lost PE would be the gained KE, so I assumed the GPE lost to falling( 1 * 10 * 5 = 50 J ) was converted to KE.
I still can't wrap my head around the process, could you please give me some pointers?
I thought the lost PE would be the gained KE, so I assumed the GPE lost to falling( 1 * 10 * 5 = 50 J ) was converted to KE.
I still can't wrap my head around the process, could you please give me some pointers?
Consider:
What's the loss in GPE (sliding down the roof and then falling)?
What's the gain in KE?
So, what's the energy lost to friction on the roof (i.e. work done against friction)?
Then it's a simple case of "Work done" = "Force" X "distance moved in the direction of that force".
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#5
Can't we find the final vertical velocity using v^2 = u^2 + 2as ?
Because we have v = 8 m/s a = 10 m/s^2 s = 5 m
Because we have v = 8 m/s a = 10 m/s^2 s = 5 m
Last edited by wasabiandyogurt; 1 month ago
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#6
(Original post by wasabiandyogurt)
Can't we find the final vertical velocity using v^2 = u^2 + 2as ?
Because we have v = 8 m/s a = 10 m/s^2 s = 5 m
Can't we find the final vertical velocity using v^2 = u^2 + 2as ?
Because we have v = 8 m/s a = 10 m/s^2 s = 5 m
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#7
(Original post by ghostwalker)
You are considering vertical motion, in which case v would be the final vertical velocity. Since it's been sliding down an inclined roof the final speed of 8m/s when it hits the ground is not vertical.
You are considering vertical motion, in which case v would be the final vertical velocity. Since it's been sliding down an inclined roof the final speed of 8m/s when it hits the ground is not vertical.
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#8
(Original post by ghostwalker)
You are considering vertical motion, in which case v would be the final vertical velocity. Since it's been sliding down an inclined roof the final speed of 8m/s when it hits the ground is not vertical.
You are considering vertical motion, in which case v would be the final vertical velocity. Since it's been sliding down an inclined roof the final speed of 8m/s when it hits the ground is not vertical.
If you have, would you please share me some perspective on how to solve it?
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#9
(Original post by wasabiandyogurt)
But the initial velocity of the vertical motion can be the final for the motion along the inclined roof right?
But the initial velocity of the vertical motion can be the final for the motion along the inclined roof right?
(Original post by wasabiandyogurt)
Did you get the answer? I have been working on this for hours.
If you have, would you please share me some perspective on how to solve it?
Did you get the answer? I have been working on this for hours.
If you have, would you please share me some perspective on how to solve it?
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#10
(Original post by ghostwalker)
I suggest you call it freefall, rather than vertical motion, since the motion once it leaves the roof is not vertical! The velocity when it leaves the roof is at an angle of 20 degrees to the horizontal.
I've not worked it through. Read the thread - I've gave a method 10 months ago!
I suggest you call it freefall, rather than vertical motion, since the motion once it leaves the roof is not vertical! The velocity when it leaves the roof is at an angle of 20 degrees to the horizontal.
I've not worked it through. Read the thread - I've gave a method 10 months ago!
Thank you so much!
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#11
(Original post by wasabiandyogurt)
Oh, that is true, I never considered that. But I didn't understand the method of the conservation of motion.
Oh, that is true, I never considered that. But I didn't understand the method of the conservation of motion.
Looking in more detail, the original question is inconsistent, so if you try and do it using suvat you will run into problems.
The change in vertical height for the freefall part would result in a vertical velocity of 10m/s if the mass started with zero vertical velocity. Since it already had some downward velocity when it started that part; when it hits the ground, the vertical velocity must be >10m/s and consequently the overall speed must be >10m/s. So, the given speed of 8m/s is impossible.
Edit: Actually, this was the OP's original issue, though I didn't realise it at the time. Any attempt to split the motion up into two parts, whether using suvat or energy considerations runs into this problem (of falling foul of the error in the question).
Last edited by ghostwalker; 1 month ago
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#12
(Original post by ghostwalker)
Not clear what you mean by the "method of the conservation of motion".
Looking in more detail, the original question is inconsistent, so if you try and do it using suvat you will run into problems.
The change in vertical height for the freefall part would result in a vertical velocity of 10m/s even if the mass started with zero vertical velocity. Since it already had some downward velocity when it started that part; when it hits the ground, the vertical velocity must be >10m/s and consequently the overall speed must be >10m/s. So, the given speed of 8m/s is impossible.
Edit: Actually, this was the OP's original issue, though I didn't realise it at the time. Any attempt to split the motion up into two parts, whether using suvat or energy considerations runs into this problem (of falling foul of the error in the question).
Not clear what you mean by the "method of the conservation of motion".
Looking in more detail, the original question is inconsistent, so if you try and do it using suvat you will run into problems.
The change in vertical height for the freefall part would result in a vertical velocity of 10m/s even if the mass started with zero vertical velocity. Since it already had some downward velocity when it started that part; when it hits the ground, the vertical velocity must be >10m/s and consequently the overall speed must be >10m/s. So, the given speed of 8m/s is impossible.
Edit: Actually, this was the OP's original issue, though I didn't realise it at the time. Any attempt to split the motion up into two parts, whether using suvat or energy considerations runs into this problem (of falling foul of the error in the question).
So, basically, the question is wrong?
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#13
(Original post by wasabiandyogurt)
Sorry, I meant 'conservation of energy'.
So, basically, the question is wrong?
Sorry, I meant 'conservation of energy'.
So, basically, the question is wrong?
Question is wrong in the sense that the information given is not consistent. It's a bit like saying a rod ABC is 4m long (and the question was worked out using that information), and AB = 3m and BC = 2m. It's an impossible situation. As long as you don't need to use the distances AB and BC, you don't have problem, but if you chose a method that requires them you'll run into trouble.
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#14
(Original post by ghostwalker)
You should be able to get to the desired answer if you use the method I suggested - not actually worked it out myself.
Question is wrong in the sense that the information given is not consistent. It's a bit like saying a rod ABC is 4m long (and the question was worked out using that information), and AB = 3m and BC = 2m. It's an impossible situation. As long as you don't need to use the distances AB and BC, you don't have problem, but if you chose a method that requires them you'll run into trouble.
You should be able to get to the desired answer if you use the method I suggested - not actually worked it out myself.
Question is wrong in the sense that the information given is not consistent. It's a bit like saying a rod ABC is 4m long (and the question was worked out using that information), and AB = 3m and BC = 2m. It's an impossible situation. As long as you don't need to use the distances AB and BC, you don't have problem, but if you chose a method that requires them you'll run into trouble.
Thank you very much! You helped a lot.
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(Original post by ghostwalker)
You should be able to get to the desired answer if you use the method I suggested - not actually worked it out myself.
Question is wrong in the sense that the information given is not consistent. It's a bit like saying a rod ABC is 4m long (and the question was worked out using that information), and AB = 3m and BC = 2m. It's an impossible situation. As long as you don't need to use the distances AB and BC, you don't have problem, but if you chose a method that requires them you'll run into trouble.
You should be able to get to the desired answer if you use the method I suggested - not actually worked it out myself.
Question is wrong in the sense that the information given is not consistent. It's a bit like saying a rod ABC is 4m long (and the question was worked out using that information), and AB = 3m and BC = 2m. It's an impossible situation. As long as you don't need to use the distances AB and BC, you don't have problem, but if you chose a method that requires them you'll run into trouble.
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#16
(Original post by Ethan7702)
I think it'll be wise to delete the thread.
I think it'll be wise to delete the thread.
The whole point of leaving threads is that they become a resource for future students.
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#17
(Original post by wasabiandyogurt)
I got the answer! But we have to assume the tile falls off the roof vertically.
Thank you very much! You helped a lot.
I got the answer! But we have to assume the tile falls off the roof vertically.
Thank you very much! You helped a lot.
What was your method then, that required the tile falls vertically?
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#18
(Original post by ghostwalker)
I've worked it through now using the method I suggested and get 9.42N; and that doesn't require the assumption of the tile falling vertically.
What was your method then, that required the tile falls vertically?
I've worked it through now using the method I suggested and get 9.42N; and that doesn't require the assumption of the tile falling vertically.
What was your method then, that required the tile falls vertically?
Then we used that for the motion on the roof. u = 0 ms^-1 and v= 6 ms^-1
We found the loss of kinetic energy using the speeds = 18 J
We found the gravitational potential change as well using (1 * 10 * 3sin 20) = 10.26 J
The we had to sum them up and divide by 3 to get the answer = 9.42 J
I feel confused however because we had to sum up the loss of energies at the end but otherwise this is what was done.
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#19
(Original post by ghostwalker)
I've worked it through now using the method I suggested and get 9.42N; and that doesn't require the assumption of the tile falling vertically.
What was your method then, that required the tile falls vertically?
I've worked it through now using the method I suggested and get 9.42N; and that doesn't require the assumption of the tile falling vertically.
What was your method then, that required the tile falls vertically?
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#20
(Original post by wasabiandyogurt)
Well, first I used v^2 = u^2 + 2as and found the initial velocity of the motion which is 6ms^-1.
Then we used that for the motion on the roof. u = 0 ms^-1 and v= 6 ms^-1
We found the loss of kinetic energy using the speeds = 18 J
We found the gravitational potential change as well using (1 * 10 * 3sin 20) = 10.26 J
The we had to sum them up and divide by 3 to get the answer = 9.42 J
I feel confused however because we had to sum up the loss of energies at the end but otherwise this is what was done.
Well, first I used v^2 = u^2 + 2as and found the initial velocity of the motion which is 6ms^-1.
Then we used that for the motion on the roof. u = 0 ms^-1 and v= 6 ms^-1
We found the loss of kinetic energy using the speeds = 18 J
We found the gravitational potential change as well using (1 * 10 * 3sin 20) = 10.26 J
The we had to sum them up and divide by 3 to get the answer = 9.42 J
I feel confused however because we had to sum up the loss of energies at the end but otherwise this is what was done.
64=u^2+2x10x5
So, u^2= -36 which is impossible - you can't just ignore the sign.
As to the method I used:
Loss in GPE
Gain in KE =
- this does not assume movement is vertical.So, work done on roof = "loss in GPE" - "gain in KE"
And this equals force x distance moved, 3F
Since m=1, put it all together to get:
and solve for F.
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