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Given: mass of girl = 35kg
Mass of trolley = 5kg
Total mass = 35+5 kg = 40 kg
Now
u=4 m/s
v=0m/s
s = 16m
Now by the third equation of motion,
v^2 - u^2 = 2*a*s
Substitute the values,
-(4)^2 =2*a*16
-16 = 2*16*a
=> a = - 0.5 m/s*s
Hence the force applied to the trolley is given by
F = (total mass) *acceleration
F = (40)*(-0.5) N
F = - 20 N
The work done on the trolley is therefore given by, W = F*s
W = - 20*16 N*m
W = - 320 J(negative work)
The work done by the girl will be given by,
W = (mass of girl) *(retardation) *s
W = 35*0.5*16 N*m
W = 280 J (positive work done)
Mass of trolley = 5kg
Total mass = 35+5 kg = 40 kg
Now
u=4 m/s
v=0m/s
s = 16m
Now by the third equation of motion,
v^2 - u^2 = 2*a*s
Substitute the values,
-(4)^2 =2*a*16
-16 = 2*16*a
=> a = - 0.5 m/s*s
Hence the force applied to the trolley is given by
F = (total mass) *acceleration
F = (40)*(-0.5) N
F = - 20 N
The work done on the trolley is therefore given by, W = F*s
W = - 20*16 N*m
W = - 320 J(negative work)
The work done by the girl will be given by,
W = (mass of girl) *(retardation) *s
W = 35*0.5*16 N*m
W = 280 J (positive work done)
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