An electric motor raises a load of 0.2kg, at a constant speed, through a vertical distance of 3.0m in 2s. if the acceleration of free fall is 10m /s , the power in W developed by the motor in raising the load is?

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Solution

$P o w e r, P = \frac{W}{t} o r \frac{E}{t} w h e r e E i s E n e r g y a n d t i s t i m e E = m a s s (m) \times g \times h e i g h t (h) = 0.2 \times 10 \times 3 = 6 J \Rightarrow P = 6 / 2 P = 3 w a t t$

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An electric motor raises a load of 0.2 kg, at a constant speed, through a vertical distance of 3.0 m in 2 s. If the acceleration of free fall is 10 $m / s^{2}$ , the power in W developed by the motor in raising the load is :

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An electric motor raises a load of 0.2kg, at a constant speed, through a vertical distance of 3.0m in 2s. if the acceleration of free fall is 10m /s , the power in W developed by the motor in raising the load is?

Power ,P=WtorEtwhere Eis Energy and t is time E=mass(m)×g×height(h) =0.2×10×3 =6 J ⇒P=6/2 P =3watt

$P o w e r, P = \frac{W}{t} o r \frac{E}{t} w h e r e E i s E n e r g y a n d t i s t i m e E = m a s s (m) \times g \times h e i g h t (h) = 0.2 \times 10 \times 3 = 6 J \Rightarrow P = 6 / 2 P = 3 w a t t$