Question

A $0.5$ HP motor is used to lift load at a constant speed of $10cm/s$. The maximum load it can lift at this speed is?
A. $370kg$
B. $530kg$
C. $185kg$
D. $555kg$

Answer 1

Hint:Power of a motor can be expressed in terms of force by which it moves multiplied by the velocity. So, here we can use the power 0.5HP (1HP= 1 Horsepower= 746 Watt) and consider the weight to be moved as force required with a constant velocity. where P is power, F is the force and v is the velocity. For this question, where mg is the load.

Complete step by step answer:
In calculus terms, we write power as the derivative of work with respect to time. So, if work is done at a faster rate, then power is higher. If work is done slower, then power is smaller. Since work is expressed as force times displacement $\left( {W = F.s} \right)$ and velocity is displacement over time (v=s/t) , power equals force times velocity: $P = F \times v$ . More power can be said when the system is both strong in force and fast in velocity.
So, in this question we are given that the motor operates at 0.5 HP.
1 HP= 1 Horsepower = 746 Watt

Let the power of motor be P and the load that it can lift at a velocity v be m.
$P = 0.5HP
\Rightarrow P = 0.5 \times 746W \\
\Rightarrow P = 373W$
$v = 10cm/s \\
\Rightarrow v= 0.1m/s$
Therefore,
$
P = mg \times v \\
\Rightarrow 373 = m \times 10 \times 0.1 \\
\therefore m = 373kg \approx 370kg \\
 $
Therefore, the correct answer is option A.

Note:In this question, the main scope of mistake is the power force relation- Power is the product of force and velocity. It's a scalar quantity. The SI unit of power is watt which is defined as “One joule of energy spent in one second.” One of the major confusion here is how we reach the power force relation, actually that relation has not to be derived every time but just memorized and used directly. Sometimes within the electrical measurements, the unit is expressed as watt-second. However, the commercial unit of power is kilowatt-hour. One kilowatt-hour is the work worn out one hour by place of work whose power is one kilowatt.