Question

Can someone help me to understand this physics problem? Thank you!

Answer 1

An engine with mass of 200 kg must be lifted 5 meters out of a truck. A normal mechanic can lift it with about 125 N of pulling force. The mechanic uses pulleys to do this. If the engine starts on the ground, how long must the ropes be on the pulley to allow the mechanic to lift the engine that distance?

Answer 2

5 meters, assuming the pulleys are inextensible. The reason is that for each distance the car moves up, so must the pulley, so a minimum length of 5 meters is required. To be even more correct, you would have to account for the pulley diameter, but I'm assuming this is a school question so I would go with 5m.

Answer 3

A small correction, I meant the rope must move, not the pulley!

Answer 4

Okay so I think I'll use W=deltaKE=Fd
but I'll need to find the change in KE to use this equation in order to find d
The change in KE should be \[\frac{ 1 }{ 2 }mv ^{2}_{f}-\frac{ 1 }{ 2 }mv ^{2}_{i}\]
the missing variable is v so I'll use this equation
\[\frac{ 1 }{ 2 }mv ^{2}_{i}+mgh _{i}=\frac{ 1 }{ 2 }mv ^{2}_{f}+mgh _{f}\]
since the engine starts on the ground it has an initial PEg of zero and assuming it stops at the end, it will have a final KE of zero simplifying it to this equation
\[\frac{ 1 }{ 2 }mv ^{2}_{i}=mgh _{f}\]
so plugging values in...
\[\frac{ 1 }{ 2 }(200kg)v ^{2}=(200kg)(9.81m/s^2)(5m)\]
solving for v gives that v=9.9m/s
so I'll plug that into the original equation
\[\frac{ 1 }{ 2 }(200kg)(9.9m/s) ^{2} -0=(125N)d\]
solving for d gives that d=78.4 and I think that would be my answer..
Do you think this looks right?

Answer 5

I think this is about machines (ie not petrol engines - no power source except the hapless, and strong, mechanic) featuring mechanical advantage, velocity ratio and efficiency, and a system of pulleys.
Can't remember the exact details, but enclosed pptx may help.
http://perendis.webs.com

Answer 6

i think you should have a look at this



you are trying to lift \( 200 \times 9.8 N\) using 125N of force, hence the pulley

but from the drawing posted how do you determine the length of the rope unless the design of the pulley is agreed first

Answer 7

you can owlet-packet this by saying that (with \(g = 10 ms^{-2}\) to make it a bit easier on eye):


Energy In = Work In:

\(mgh = F \times l\)

\(\implies 200 \times 10 \times 5 = 125 \times l \implies l = 80m\)

that is the distance over which the 125N force must act to raise the engine 5 m upwards; but whether or not it is the length of the rope is another [related] question.

Answer 8

@osprey

just opened your attachment and see we are ad idem

Answer 9

@IrishBoy123 My knowledge of Latin is as bad as my physics .... ? "ad idem" . ad = to (ad nauseam), "idem", though ?

Answer 10

īdem m ‎(feminine eadem, neuter idem)

= the same