
If the power of a motor is $40{\text{kW}}$, at what speed can it raise a load $20,000{\text{N}}$?
$1)0.2{\text{m}}{{\text{s}}^{ - 1}}$
$2)2{\text{m}}{{\text{s}}^{ - 1}}$
$3)20{\text{m}}{{\text{s}}^{ - 1}}$
$4)$ None of the above
Answer
502.2k+ views
Hint: We know that power is equal to the rate of doing work. It is the measure of work done. In the question, we are given the value of power and the force applied and we need to find the value of the speed at which the motor can raise the load. We need to define power in terms of force and velocity.
Complete step by step solution:
Power is a unit to measure the work done. It can be expressed in terms of work done per unit time. It can also be explained in terms of voltage, energy, force etc. It is defined as the amount of work done in converting the energy from one form to another in a given time.
Therefore we have
$P = \dfrac{W}{t}$
Where $W$is the work done and $t$ is the time taken.
We know that work done is equal to the product of the force applied and the distance covered.
Therefore we have $W = F \times S$
Where $F$ is the force applied and $S$ is the displacement
Therefore power can be written as
$P = \dfrac{{F \times S}}{t}$
Now we know that the distance covered per unit time is known as velocity. Therefore we have
$P = F \times v$
Where $v$ is the velocity of the motor.
Hence substituting the values we get
$40000 = 20000 \times v$
$ \Rightarrow v = 2{\text{m}}{{\text{s}}^{ - 1}}$
Hence option 2) is the correct option.
Note: Note that when we are describing the power in terms of current and electricity, we express the power in terms of voltage and current but when we describe the power for rotational motion we express in terms of the torque and angular velocity.
Complete step by step solution:
Power is a unit to measure the work done. It can be expressed in terms of work done per unit time. It can also be explained in terms of voltage, energy, force etc. It is defined as the amount of work done in converting the energy from one form to another in a given time.
Therefore we have
$P = \dfrac{W}{t}$
Where $W$is the work done and $t$ is the time taken.
We know that work done is equal to the product of the force applied and the distance covered.
Therefore we have $W = F \times S$
Where $F$ is the force applied and $S$ is the displacement
Therefore power can be written as
$P = \dfrac{{F \times S}}{t}$
Now we know that the distance covered per unit time is known as velocity. Therefore we have
$P = F \times v$
Where $v$ is the velocity of the motor.
Hence substituting the values we get
$40000 = 20000 \times v$
$ \Rightarrow v = 2{\text{m}}{{\text{s}}^{ - 1}}$
Hence option 2) is the correct option.
Note: Note that when we are describing the power in terms of current and electricity, we express the power in terms of voltage and current but when we describe the power for rotational motion we express in terms of the torque and angular velocity.
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