How much work is required to raise a 0.1 kg block to a height of 2 m and simultaneously give it a constant velocity of 3 m/s? b) What is the change in its potential energy? c) What is the power output of the upward force in watts?
(a)W=mgh+12mv2W=mgh+\frac{1}{2}mv^2W=mgh+21mv2
=(0.1×9.8×2)+(12×0.1×32)=(0.1\times9.8\times2)+(\frac{1}{2}\times0.1\times3^2)=(0.1×9.8×2)+(21×0.1×32)
=2.41J=2.41J=2.41J
(b) △P=P2−P1 \triangle P=P_2-P_1 \space△P=P2−P1 but P1=0P_1=0P1=0
△P=mgh=0.1×9.8×2=1.96J\triangle P=mgh=0.1\times9.8 \times2=1.96J△P=mgh=0.1×9.8×2=1.96J
(c) P=F×V=0.1×9.8×3=2.94WP=F\times V=0.1\times 9.8\times3=2.94WP=F×V=0.1×9.8×3=2.94W
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