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Zorluk: KolayWork, Energy and Power

A mountain climber of mass 60 kg60\text{ kg} climbs a vertical height of 15 m15\text{ m} in a time of 30 s30\text{ s}. Taking the acceleration due to gravity g=10 m s−2g = 10\text{ m s}^{-2}, calculate the average power expended by the climber in Watts.

Cevap: 300 W

Cevap

The average power expended by the climber is 300 W300\text{ W}.
The work done in lifting a mass mm through vertical height hh is given by W=mgh=60×10×15=9000 JW = mgh = 60 \times 10 \times 15 = 9000\text{ J}. The average power is the rate of doing work, P=Wt=9000 J30 s=300 WP = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}.

Adım Adım Çözüm

1
Identify the given values and formula for work done against gravity.
Mass m=60 kgm = 60\text{ kg}, vertical displacement h=15 mh = 15\text{ m}, acceleration due to gravity g=10 m s−2g = 10\text{ m s}^{-2}, and time t=30 st = 30\text{ s}. Work done formula: W=mghW = mgh.
The climber works against gravity to increase their potential energy by an amount equal to mghmgh.
2
Calculate the total work done.
W=60 kg×10 m s−2×15 m=9000 JW = 60\text{ kg} \times 10\text{ m s}^{-2} \times 15\text{ m} = 9000\text{ J}.
Multiplying force (mgmg) by vertical distance (hh) yields work done in Joules.
3
Calculate the average power.
P=Wt=9000 J30 s=300 WP = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}.
Power is defined as work done divided by the time interval (P=WtP = \frac{W}{t}).

Anahtar Kavram

Power as the rate of doing work against gravity
Tahmini Süre:45s
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Work, Energy and Power

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