As per the question the mass of the boy is 50 kg.
The boy sits on a chair.
We are asked to calculate the force exerted by the boy on the chair at sea level.
The force exerted by boy on the chair while sitting on it is nothing else except the force of gravity of earth i.e the weight of the body .The direction of that force is vertically downward.
At sea level the acceleration due to gravity g = 9.8 m/s^2
Hence the weight of the boy 
[m is the mass of the body]
we have m = 50 kg.
Hence w = 50 kg ×9.8 m/s^2
=490 N kg m/s^2
= 490 N
Here newton [N] is the unit of force.
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Answer:
i think its C hope this is rhight have a good day
Explanation:
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
