
A truck weighing $1000\,kg$ changes its speed from $36km\,{h^{ - 1}}$ to $72km\,{h^{ - 1}}$ in $2{\text{ minutes}}$ . Calculate: $\left( i \right)$ the work done by the engine, and $\left( {ii} \right)$ its power. $\left( {g = 10m\,{s^{ - 2}}} \right)$
Answer
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Hint: For solving this question we have to first find the work done and for this, we will use the formula $\dfrac{{m\left( {{v^2} - {u^2}} \right)}}{2}$ and by substituting the values we will get the solution. Similarly, we know that power is defined as the work done per unit of time. So by using this we will get the value for the power.
Formula used:
Work is done,
$W = \dfrac{{m\left( {{v^2} - {u^2}} \right)}}{2}$
Here, $m$ will be the mass
$v$ , will be the final velocity
$u$ , will be the initial velocity
Power,
$P = \dfrac{W}{t}$
Here, $P$ will be the power
$W$ , will be the work done
$t$ , will be the time
Complete step by step answer:
So we have the values given to us as,
$ \Rightarrow v = 72km\,{h^{ - 1}}$
Converting the above unit into meter per second
$ \Rightarrow 72 \times \dfrac{5}{{18}}$
And on solving it, we get
$ \Rightarrow v = 20m/\sec $
Similarly, $u = 36km\,{h^{ - 1}}$
Converting the above unit into meter per second
$ \Rightarrow 36 \times \dfrac{5}{{18}}$
And on solving it, we get
$ \Rightarrow u = 10m/\sec $
We have time given as $2{\text{ minutes}}$ . So on converting it into seconds, we get
$ \Rightarrow t = 60 \times 2 = 120\sec $
Now by using the formula of work done and substituting the values, we get
$ \Rightarrow W = \dfrac{{1000\left( {{{20}^2} - {{10}^2}} \right)}}{2}$
On solving the above equation, we get the equation as
$ \Rightarrow W = 150 \times {10^3}J$
And on solving it, we get
$ \Rightarrow W = 150kJ$
Hence, the work done by the engine is equal to $150kJ$ .
Now by using the formula of power, and substituting the values, we get
$ \Rightarrow P = \dfrac{{150 \times {{10}^3}}}{{120}}$
Now on solving the above equation, we get the value of power as
$ \Rightarrow P = 1.25 \times {10^3}w$
And it can also be written as
$ \Rightarrow P = 1.25kw$
Hence, its power will be $1.25kw$ .
Note: Work done is a measure of consuming energy or it is the capacity of doing a certain work and that also means the same . Work is something which we do which consumes or produces energy or there will be a net increase or decrease of energy.
Formula used:
Work is done,
$W = \dfrac{{m\left( {{v^2} - {u^2}} \right)}}{2}$
Here, $m$ will be the mass
$v$ , will be the final velocity
$u$ , will be the initial velocity
Power,
$P = \dfrac{W}{t}$
Here, $P$ will be the power
$W$ , will be the work done
$t$ , will be the time
Complete step by step answer:
So we have the values given to us as,
$ \Rightarrow v = 72km\,{h^{ - 1}}$
Converting the above unit into meter per second
$ \Rightarrow 72 \times \dfrac{5}{{18}}$
And on solving it, we get
$ \Rightarrow v = 20m/\sec $
Similarly, $u = 36km\,{h^{ - 1}}$
Converting the above unit into meter per second
$ \Rightarrow 36 \times \dfrac{5}{{18}}$
And on solving it, we get
$ \Rightarrow u = 10m/\sec $
We have time given as $2{\text{ minutes}}$ . So on converting it into seconds, we get
$ \Rightarrow t = 60 \times 2 = 120\sec $
Now by using the formula of work done and substituting the values, we get
$ \Rightarrow W = \dfrac{{1000\left( {{{20}^2} - {{10}^2}} \right)}}{2}$
On solving the above equation, we get the equation as
$ \Rightarrow W = 150 \times {10^3}J$
And on solving it, we get
$ \Rightarrow W = 150kJ$
Hence, the work done by the engine is equal to $150kJ$ .
Now by using the formula of power, and substituting the values, we get
$ \Rightarrow P = \dfrac{{150 \times {{10}^3}}}{{120}}$
Now on solving the above equation, we get the value of power as
$ \Rightarrow P = 1.25 \times {10^3}w$
And it can also be written as
$ \Rightarrow P = 1.25kw$
Hence, its power will be $1.25kw$ .
Note: Work done is a measure of consuming energy or it is the capacity of doing a certain work and that also means the same . Work is something which we do which consumes or produces energy or there will be a net increase or decrease of energy.
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