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Having major problems with this question MECHANICS

yonewt

hey can someone please provide a comprehensible outline as to how to approach this question 5b)

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Reply 1

teachercol

Total Power = work done against friction per s + total gain in pe per s.

That should get you started

Reply 2

Arbolus

You can find the work required to lift one package up the slope. Then convert 50000 packages per hour into packages per second and multiply the work by that rate to find the power involved in lifting.

Reply 3

yonewt

Original post by Arbolus

how do i get the distance to find the work and if i dont need the distance , how do i find the work without it?

Reply 4

Stonebridge

Original post by yonewt

how do i get the distance to find the work and if i dont need the distance , how do i find the work without it?

You need the vertical height the packages are lifted.
The distance along the slope is given (hypotenuse) and so is the sine of the angle. From this you can find the height (opposite side).

Reply 5

Mort89

Original post by yonewt

how do i get the distance to find the work and if i dont need the distance , how do i find the work without it?

You don't need the work you need the power.

power=work/time, velocity=distance/time,

therefore just use the velocity instead of the distance

Reply 6

yonewt

Original post by Stonebridge

You need the vertical height the packages are lifted.
The distance along the slope is given (hypotenuse) and so is the sine of the angle. From this you can find the height (opposite side).

no it isnt it is only the velocity 75ms^-1

Reply 7

Stonebridge

Original post by yonewt

no it isnt it is only the velocity 75ms^-1

Err, yes it is.
Look at the question again.
It says

- the packages are moved 75m up a slope
- there are 50,000 packages moved per hour

(edited 10 years ago)

Reply 8

Mort89

Question's slightly ambiguous. It could be 50000 packages per hour are moved up a slope of length 75m, or it could be 50000 packages are move up an arbitrary length slope at a speed of 75m per hour, but either way it's the same answer.

Reply 9

Stonebridge

Original post by Mort89

Agreed it's poorly worded.

A better sentence structure would have been

"...50,000 packages per hour, 75m up a slope with inclination..."

Reply 10

yonewt

Original post by Stonebridge

Err, yes it is.
Look at the question again.
It says

- the packages are moved 75m up a slope
- there are 50,000 packages moved per hour

i know im asking a lot here but could you give me bit of an outline or layer of what i should .......... i am REALLY confused

Reply 11

Stonebridge

Original post by yonewt

i know im asking a lot here but could you give me bit of an outline or layer of what i should .......... i am REALLY confused

Find the vertical height h each package is lifted.
The distance up the slope (hypotenuse) is 75m
The sine of the angle is given so you can find this from sin = opposite / hypotenuse
It's the opposite side you need to find. That's the vertical height.

Then find the potential energy the package gains from mgh (h you just found and mg is the weight of a package which is given)
Then work out how much energy for 50,000 packages (you just worked it out for one)
Then find the power required to do this in 1 hour. (power is Energy per second. You have worked out energy per hour)
What do you get?

There's one more step after this.

(edited 10 years ago)

Reply 12

yonewt

Original post by Stonebridge

Find the vertical height h each package is lifted.
The distance up the slope (hypotenuse) is 75m
The sine of the angle is given so you can find this from ysin = opposite / hypotenuse
It's the opposite side you need to find. That's the vertical height.

Then find the potential energy the package gains from mgh (h you just found and mg is the weight of a package which is given)
Then work out how much energy for 50,000 packages (you just worked it out for one)
Then find the power required to do this in 1 hour. (power is Energy per second. You have worked out energy per hour)
What do you get?

There's one more step after this.

i got 30130.56 J/s after working out energy per second what do i do know

Reply 13

Stonebridge

Original post by yonewt

i got 30130.56 J/s after working out energy per second what do i do know

You need to show how you got that value.
What was the energy for 1 packet (mgh)
For 50,000 packets (50,000 x mgh)
Then this divided by the number of seconds in an hour.

There's a mistake somewhere. You need to show working so we can find it.