Calculate Work Done: 40kg Stone, 30o Incline, 100m Height

[lemon]

1. A 40kg stone is pulled slowly up a smooth ramp inclined at 30^o to the horizontal. It stops at the top, at a height of 100m above its initial height. Calculate the work done by:
a) Using 'work done = force x distance moved in direction of force'.
b) Using conservation of energy



2. F=ma
F-F_gravity=mgsin(theta)
Work done=Fscos(theta)



3. Weight = 400N
F-400=400sin(30)=600N
Work done = 600 x 100cos(30) = 52kJ (2sf)


Not sure this is correct
please advise?

 

[tiny-tim]

Hi lemon!

F-400=400sin(30)=600N

No … where did the " - 400 " come from?

 

[lemon]

weight - mg = 40kg x 10m/s

 

[lemon]

Right. Sorry. To calculate the F_net.
F_net=400sin(30) = 200N
Work done = 200N x 100m = 20kJ

 

[tiny-tim]

F yes , W no .

'work done = force x distance moved in direction of force'

so the "x distance" isn't 100.

 

[lemon]

ahh!
sin(theta)=opposite/hypotenuse
100/sin(30) = 200m

W=200N x 200m =4000J
or 4kJ

 

[tiny-tim]

That's it!

btw, you might like to try this, just to see how it all fits together …

you've assumed that the force is applied the obvious way, by a rope parallel to the slope,

but suppose it is applied by a rope at an angle φ to the slope (φ + 30º to the horizontal) …

what is F then, and what is W? ​

And of course try b) (calculate the work done using conservation of energy)

 

[lemon]

So for b) calculate the work done using conservation of energy

Total amount of energy always stays the same, as energy cannot be created nor destroyed, just transferred to other forms.

We have gained height by inputting energy to pull the block up the slope.
So we have gone from kinetic energy to gravitational potential energy. We have calculated that the work done is 4kJ. So:
KE lost = PE gained
4kJ = mgh

I think I'm lost

 

[tiny-tim]

So for b) calculate the work done using conservation of energy

Total amount of energy always stays the same, as energy cannot be created nor destroyed, just transferred to other forms.

We have gained height by inputting energy to pull the block up the slope.

You can assume the KE is zero all the time …

1. A 40kg stone is pulled slowly …

So the work done is just the difference in PE, which is … ?

 

[lemon]

mgh - 4kJ
(40kg x 10m/s x 100m) - 4kJ
= 36000j
36kJ

 

[tiny-tim]

mgh - 4kJ
(40kg x 10m/s x 100m) - 4kJ
= 36000j
36kJ

No, the work done is just the difference in PE

(oh, and I've just noticed you miscounted the 0s in …)

W=200N x 200m =4000J

 

[lemon]

How can there be a difference in PE when there is no PE at the bottom of the slope?

 

[tiny-tim]

There isn't no PE … the PE is 0.

So the difference is … ? ​

 

[lemon]

40kJ - 0?

 

[tiny-tim]

40kJ - 0?

yes, the work done is just mgh.

"Work done" is another name for the difference in PE (in a conservative field).

ok, your 40kJ is now the same as you should have got under part a).

 

[lemon]

Man! These questions are brain twisters. I look at them and think they look straight forward, but man! I'm really glad I decided to come on here and get help with these or I don't think I would hardly have understood at all. Is it just me?

Last part:
It is now dropped from the top of the building descending vertically in free fall. Calculate its speed just before it hits the ground. What happens to its kinetic energy?

This is PE to KE
mgh = 1/2mv²
40kJ = 1/2x40xv²
root2000 = 20root5 or 44.7m/s

Kinetic energy is then transferred into any deformation of the ground/stone, and into sound and heat. Is that what the question is looking for?

 

[tiny-tim]

Yes, that's fine.

 

[lemon]

Thanks again tT