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Question: A truck weighing 1000kg f changes its speed from \(36km{h^{ - 1}}\) to \(72km{h^{ - 1}}\) in 2 minut...

A truck weighing 1000kg f changes its speed from 36kmh−136km{h^{ - 1}} to 72kmh−172km{h^{ - 1}} in 2 minutes, (g=10ms−2)\left( {g = 10m{s^{ - 2}}} \right). Calculate the work done by the engine.
A)1.5×105J B)1×105J C)7.2×105J D)0J  A) 1.5 \times {10^5}J \\\ B) 1 \times {10^5}J \\\ C) 7.2 \times {10^5}J \\\ D) 0 J \\\

Explanation

Solution

Hint: In order to solve this question, firstly we will convert the given units in kilometer per hour to meter per second. Then we will use the formula of work done i.e. W=f×sW = f \times s where f=m×af = m \times a, so that we get our desired result.

Complete step-by-step answer:
Here we are given that –

Mass of the truck = 1000kg
Firstly we will convert the given speed in kilometer per hour to meter per second.

Initial speed (u)=36km per hour = 36×1000m3600s=10m per second = 36km{\text{ per hour = }}\dfrac{{36 \times 1000m}}{{3600s}} = 10m{\text{ per second}}
Increased speed (v)=72km per hour = 72×1000m3600s=20m per second = 72km{\text{ per hour = }}\dfrac{{72 \times 1000m}}{{3600s}} = 20m{\text{ per second}}
t=2min⁡=2×60=120t = 2\min = 2 \times 60 = 120 seconds

As we know that,

We know that work done is equal to the product of force and displacement.
Also we know that force is equal to the product of mass and acceleration.
Now using the formula of acceleration i.e.
a=v−uta = \dfrac{{v - u}}{t}.

Substituting the values of v=20m/s, u=10m/s and t=120 seconds in the above formula,
We get-

a=20−10120=112m per s2a = \dfrac{{20 - 10}}{{120}} = \dfrac{1}{{12}}m{\text{ per }}{{\text{s}}^2}

Now using the newton’s third law of motion i.e. v2−u2=2as{v^2} - {u^2} = 2as,
We get-

(v−u)(v+u)=2×(v−u)t×s\left( {v - u} \right)\left( {v + u} \right) = 2 \times \dfrac{{\left( {v - u} \right)}}{t} \times s
wheres=12(v+u)t=12(20+10)×120=1800ms = \dfrac{1}{2}\left( {v + u} \right)t = \dfrac{1}{2}\left( {20 + 10} \right) \times 120 = 1800m

Therefore, we get-

W=m×a×sW = m \times a \times s

Now substituting the values of m=1000kg, a=112m per s2a = \dfrac{1}{{12}}m{\text{ per }}{{\text{s}}^2} and s=1800m
W=1000×1×180012W = \dfrac{{1000 \times 1 \times 1800}}{{12}}
W=1.5×105JW = 1.5 \times {10^5}J

Therefore, we conclude that the work done by the engine, W=1.5×105JW = 1.5 \times {10^5}J.
Hence, option A is correct.

Note- While solving this question, we can also use another method i.e. using the concept of mechanical energy we can take work done equals to kinetic energy by converting the given units in kilometer per hour to meter per second and using the given mass and difference of the speed to get the required result.