Explanation
Given that, speed of particle is becoming 3 times of its initial value while moving. ie $$v'=3v$$
As we know $$K.E = \dfrac{1}{2}$$ $$ mv ^{2}$$ ..........(1)
After velocity tripled $$K.E' = \dfrac{1}{2}$$ $$ m(3v) ^{2}=\dfrac92mv^2$$...........(2)
From equation 1 and 2
$$K.E' = 9 \times K.E$$
Hint: Work done is the area under F-x graph.
Correct Option: B
$$\textbf{Step1: Calculation of work done}$$
W = Area of $$ABNM$$ + Area of $$CDEN$$- Area of $$EFGH$$ + Area of $$HIJ$$
(Area of $$EFGH$$ is taken negative as the force is in negative direction)
$$ \Rightarrow W = \left( {1 \times 10} \right) + \left( {1 \times 5} \right) - \left( {1 \times 5} \right) + \left( {\dfrac{1}{2} \times 10 \times 1} \right) = 15J$$
So, the work done in displacing the body from x=1 m to x=5 m will be $$15J$$ .
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