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#1
this guy cycles up a slope (combined mass of bike and guy is 80kg) inclined at arcsinx=1/7 to the horizontal. Resistance against motion = 40N and he travels at a constant speed of 6m/s.

a) find the power exerted by the man
b)If the man suddenly increases his workrate by 240 W, find the magnitude of his acceleration up the slope at this instant.

I have done a) and got 912W but dont know how to do b)
the answer to b) is 0.5 m/s^-2

0
13 years ago
#2
this guy cycles up a slope (combined mass of bike and guy is 80kg) inclined at arcsinx=1/7 to the horizontal. Resistance against motion = 40N and he travels at a constant speed of 6m/s.

a) find the power exerted by the man
Equation of motion parallel to slope
F - 40 - 80gsinx = ma
F - 40 - 80gsinx = 0
F = 40 + 80gsinx
F = 40 + 80[9.8][1/7]
F = 152N

Power = Force x velocity
Power = 152 x 6

Power = 912W

b)If the man suddenly increases his workrate by 240 W, find the magnitude of his acceleration up the slope at this instant.

I have done a) and got 912W but dont know how to do b)
the answer to b) is 0.5 m/s^-2

Equation of motion parallel to slope
F - 40 - 80gsinx = ma
F - 40 - 80[9.8][1/7] = 80a
F - 40 - 112 = 80a

Power = Force x velocity
1152 = Fv
1152 = 6F
F = 1152/6 = 192N

a = [F - 40 - 112]/80
a = [192 - 40 - 112]/80
a = 40/80
a = 0.5ms-2
0
#3
Equation of motion parallel to slope
F - 40 - 80gsinx = ma
F - 40 - 80[9.8][1/7] = 80a
F - 40 - 112 = 80a

Power = Force x velocity
1152 = Fv
1152 = 6F
F = 1152/6 = 192N

a = [F - 40 - 112]/80
a = [192 - 40 - 112]/80
a = 40/80
a = 0.5ms-2[/QUOTE]

ok thanx. the thing that confuses me is why velocity remains 6ms-1 when acceleration has increased
0
13 years ago
#4
The key is the phrase "at this instant."
He has just, at that moment, increased his power, so meaning that his acceleration will be greater, but there hasn't been time for his velocity to actually have increased yet, if you get what I mean.
0
#5
(Original post by Narin)
The key is the phrase "at this instant."
He has just, at that moment, increased his power, so meaning that his acceleration will be greater, but there hasn't been time for his velocity to actually have increased yet, if you get what I mean.
yeah thanx alot i know what u mean. well explained lol
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