Energy Question (1 Viewer)

[Husayn]

A load with a mass of 2000 kg is hauled up an incline of 1 in 100. The frictional resistance opposing the motion up the plane is constant at 300 N. What energy is needed to get the load a distance of 4m up the slope?

 

[fakingtheday]

First thing we have to do is work out the angle of inclination.

Tan (*) = 1/100
(*) = 0.01
Tan-1 = 0.57 degrees

Now break the 20kn force into forces parallel and perpendicular to the plane ( i didn't do the perp. isn't needed):

Sin 0.57 = x/20 000
x = 199 N

Ok so now we have a component of weight down the plan and a constant frictional force. Therefore the total force to be overcome:

Force (total) = 199 + 300
F = 499 N

Now to find the energy:

Energy = Force x Distance
E = 499 x 4
E = 1996 nM (Joules)
THerefore total energy required = 1.996 KJ


Did i get that one right?

 

[Husayn]

No, totally wrong,

It's 1.98 kJ.

 

[Husayn]

OK another one:

A car of mass 1200 kg starts from rest and reaches a speed of 20 m/s after travelling 250m along a straight road. If the driving force is constant, what is its value?

 

[S13WPN]

OK another one:

A car of mass 1200 kg starts from rest and reaches a speed of 20 m/s after travelling 250m along a straight road. If the driving force is constant, what is its value?

= 1500

how'd i go??

Edit: or maybe it's 6000?

 

[fakingtheday]

How'd you do that? I got 240N

It's moving so it has Kinetic Energy = 1/2 x mass x velocity(2)
So:
Determin the velocity. Start with 0, finish with 20 therefore (20 + 0)/2 = 10m/s

KE = 1/2mv(2)
KE = 1/2 x 1200 x 10(2)
KE = 60 000 J
= 60 KJ

Substitute into the formula:

Work = Force x Distance
60 000 = Driving Force x 250
Driving force = 240N

 

[S13WPN]

alright i see what you've done here lol, i was just looking over it before i read ur post... it seemed a bit wrong but i think i stumbled onto the work part of it

6000

but yeh i used the wrong formula or somethin

 

[fakingtheday]

So you know how to do it now? my mechanics has died in the ass the last few weeks. Did you get the 240?

 

[S13WPN]

i can see how you got the answer... i forgot all about that 1/2mv(2) formula so thanks for that, but my physics knowledge isnt great either

 

[fakingtheday]

I'm making one up:

A cyclist of mass 110KG is travelling at a constant velocity of 5m/s along a surface that has no frictional reisistance and comes to a slope of 20*. If he stops pedalling how far up the slope will he roll before he stops completely?

 

[Husayn]

The answer is 960 N.

I think because the force is constant you don't need an average, so I did:

.5 x 1200 x 20^2 = 240,000

240,000 = F x 250

F = 960 N

There are a few more examples like this which I'll post later.

-----

m = 110 kg
v = 5
theta = 20

KE = .5 x 110 x 5^2 = 1375

No frictional resistance, so then all that's opposing him is his weight? 1100 x sin 20 I assume? I'm just guessing, so:

1375 = 1100Sin 20 x d

d = 1375/1100 sin 20

No calculator at hand so I'll leave as is. How far off was I?

 

[fakingtheday]

Correct.

I'm pretty sure you have to take an average because there is acceleration between rest and 20 m/s. Although i remember arguing about it with my teacher once. What book was that one from?

 

[Husayn]

Engineering Science by Bolton.

 

[PimpPyro]

I got the same as you for the slope question!

I got same answer for the slope question faking made up:
3.6547555002038590652909430167106m up the slope it ends up being.

 

[fakingtheday]

I'm pretty sure my question is one of the harder types we would get.

 

[PimpPyro]

Probably I would say the hardest we would ever get.

It is pretty hard, I mean usually the energy questions are about lifting a load or something simple.

 

[fakingtheday]

It would sort out probably 97-97% of the cohort anyway.

 

[PimpPyro]

Too much BS!

Yeh I wish they gave harder mechanics questions, some of them are very weak! Too much BSing about societal and environmental BS, not enough calculations and theoretical work!