Power On An Inclined Plane

Question

A hoist motor drags a 100 kg load up a smooth incline angled at 30 degrees to the horizontal, moving it at a steady 0.5 m/s along the slope. What power output is required from the motor?

RankGuru · Accepted Answer

Pulling a load up a frictionless incline at constant speed requires a force equal to the component of gravity along the slope, $F = mg\sin	heta$, because there is no acceleration and no friction. The power needed is then this force multiplied by the speed along the incline, $P = Fv$. With $m = 100	ext{ kg}$, $g = 10	ext{ m/s}^2$, and $\sin 30^\circ = 0.5$, the force is $100 	imes 10 	imes 0.5 = 500	ext{ N}$. Multiplying by $v = 0.5	ext{ m/s}$ gives $P = 500 	imes 0.5 = 250	ext{ W}$. The option 500 W omits the velocity factor and reports the force. The option 125 W mistakenly uses $\sin 30^\circ$ twice. The option 1000 W ignores the sine component, treating the slope as vertical. The decomposition of gravity into components along and perpendicular to the slope is central here: the perpendicular component is cancelled by the normal reaction and does no work, while only the slope-parallel component resists the upward motion. Because the incline is smooth, no extra power is spent overcoming friction, so the motor's entire output is converted into gravitational potential energy of the load. This combines the NCERT incline analysis with the power-velocity relation. As a check, only the vertical lift rate $v\sin	heta$ does gravitational work, and $mg(v\sin	heta) = 100 	imes 10 	imes 0.25 = 250	ext{ W}$ agrees.

Power On An Inclined Plane

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