need help on phys qns...
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this concept i still nt very sure yet...someone clarify it for me? thx...
there is a man who is moving a box up a slope 5m long, 4m tall and 3m for the length of the base. he exerts a constant force of 30N while there is a constant frictional force of 10N.
a) what is the work done by the man?
b) what is the work done by the force he exerts?
c) what is the work done against friction?
maybe a and b will be the same. i dunno tt part totally. lol. thx! -
i THINK a and b same because constant force means the resultant force exert by the man should be the same
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i'm rusty. but ....
a) considering only force exerted by man, work done = 30N * 5m
b) Asuming gravitional force = 10m^2/s
Resultant work done = (30N - 10N - 8kg*10*Cos30) * 5m
c) work to overcome friction only = -10N * 5m -
work done = change in kinetic energy + change in potential energy
part of the weight of the block acts on it which opposes motion upwards.
(draw vector triangle, spliting weight of block into 2 components. one perpendicular to plane and one parallel to plane, both from the centre of gravity of mass)
to find that weight,
hey boy! i think force provided is not enough to move the mass up the slope!
partial weight + friction > 30N
did u set this question yourself or found it somewhere -
yea i did.cuz its these types of qns that i dont really know. but i thought according to f=ma, practically any lousy force greater than the mass of the object itself can already give it an acceleration, hence it can be moved?
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always remember that work done = force * distance moved in the direction of the force
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a) considering only force exerted by man, work done = 30N * 5m
b) Asuming gravitional force = 10m^2/s
Resultant work done = (30N - 10N - 8kg*10*Cos30) * 5m
@ (a)
if there is friction...work done by man is still the force that he exerts(30N) * distance? dun need minus frictional force right?
@(b)
i dont understand
the work done by force refer to the 30N force..why got cos 30 and all that pop out..lol -
Originally posted by eagle:
no wor, he got produce heat and sound etc to over come friction. its work done also mah, it still is a transfer of enegy. whether the energy transfer is useful or not is another issue
always remember that work done = force * distance moved [b]in the direction of the force[/b] -
that is why it is frictional forceOriginally posted by hisoka:no wor, he got produce heat and sound etc to over come friction. its work done also mah, it still is a transfer of enegy. whether the energy transfer is useful or not is another issue
Your work done against friction is converted into heat and sound and etc. Don't need to take into account any heat or sound things anymore
Total WD = force * blah blah blah
This force includes your frictional force as well as the force required to move the object -
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I need to rush to school liaoOriginally posted by ^tamago^:
u help me jaga jaga hor -
i still dont know this question...someone help me
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okOriginally posted by absol:i still dont know this question...someone help me
a) if moved 5m
Ans is 30N * 5m = 150 J
b) is same as a)
c) work done against friction is 10N * 5m = 50J -
but...but...but...but...~ i also went out wor~Originally posted by eagle:I need to rush to school liao
u help me jaga jaga hor