
1000 N force is required to lift a hook and 10000 N force is required to lift a load slowly. Find power required to lift hook with load with speed $v = 0.5m/s$
A. $5 kW$
B. $5.5 kW$
C. $1.5 kW$
D. $4.5 kW$
Answer
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Hint: The power required to lift hook with load is calculated on dividing work done in lifting the load by time taken. The work done is the product of total force and displacement and total force is the sum of force required to lift the hook and force required to lift the load. Displacement divided by time gives us the velocity or speed.
Complete step by step solution:
A force is defined as the physical cause which changes or tends to change either the size or shape of the body or the state of rest or the state of motion of the body. A force ${F_1} = 1000N$ is required to lift a hook and a force ${F_2} = 10000N$ is required to lift a load. Let F be the force required to lift the hook when it is attached to the load.$F = {F_1} + {F_2}$ i.e., $1000 + 10000 = 11000N$.
The rate of doing work is called power and the power spent by a source is defined as the amount of work done per second by the source or it is numerically equal to the work done by the source (hook +load) in one second. Let power be denoted by $P$.
Thus, $P = \dfrac{W}{t}$ where $W$ is work done and $t$ be the time taken by it.
If a constant force $F$ acts on a body and it displaces the body by a distance $S$ in the direction of force in time $t$, then work done $W = F \times S$
Now, we substitute the value of W in the above equation i.e., $P = \dfrac{W}{t}$
$P = \dfrac{{F \times S}}{t}$
$\Rightarrow P = F \times \left( {\dfrac{S}{t}} \right)$
$\Rightarrow P = F \times v$ as $v = \dfrac{S}{t}$ and v is the velocity.
$\Rightarrow P = 11000 \times 0.5$ as $v = 0.5m/s$
$\therefore P = 5500W$ or $5.5kW\left[ {1W = 0.001kW} \right]$
Therefore, option B is correct.
Note:We should remember that the term force used in the formula of power is the total force i.e., the sum of force required to lift the hook and load separately. Since the value of speed is given in the question, we have to replace distance or displacement and time with the speed as speed is the ratio of distance and time.
Complete step by step solution:
A force is defined as the physical cause which changes or tends to change either the size or shape of the body or the state of rest or the state of motion of the body. A force ${F_1} = 1000N$ is required to lift a hook and a force ${F_2} = 10000N$ is required to lift a load. Let F be the force required to lift the hook when it is attached to the load.$F = {F_1} + {F_2}$ i.e., $1000 + 10000 = 11000N$.
The rate of doing work is called power and the power spent by a source is defined as the amount of work done per second by the source or it is numerically equal to the work done by the source (hook +load) in one second. Let power be denoted by $P$.
Thus, $P = \dfrac{W}{t}$ where $W$ is work done and $t$ be the time taken by it.
If a constant force $F$ acts on a body and it displaces the body by a distance $S$ in the direction of force in time $t$, then work done $W = F \times S$
Now, we substitute the value of W in the above equation i.e., $P = \dfrac{W}{t}$
$P = \dfrac{{F \times S}}{t}$
$\Rightarrow P = F \times \left( {\dfrac{S}{t}} \right)$
$\Rightarrow P = F \times v$ as $v = \dfrac{S}{t}$ and v is the velocity.
$\Rightarrow P = 11000 \times 0.5$ as $v = 0.5m/s$
$\therefore P = 5500W$ or $5.5kW\left[ {1W = 0.001kW} \right]$
Therefore, option B is correct.
Note:We should remember that the term force used in the formula of power is the total force i.e., the sum of force required to lift the hook and load separately. Since the value of speed is given in the question, we have to replace distance or displacement and time with the speed as speed is the ratio of distance and time.
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