Question

step wise answer only​

Answer 1

By work - energy theorem, we know that,

$\longrightarrow\sf{K_f-K_i=Work\ done}$

But since the particle moves from rest, $\sf{K_i=0.}$ Then,

$\longrightarrow\sf{K_f=Work\ done}$

That is, the kinetic energy of the particle after travelling 3 metres is the work done to travel this 3 metres, which is given by the area under the graph.

Hence the kinetic energy,

$\longrightarrow\sf{K=Area\ of\ OABCD}$

$\longrightarrow\sf{K=Area\ of\ OABE+Area\ of\ BCDE}$

$\longrightarrow\sf{K=OA\cdot AB+\dfrac{1}{2}\cdot DE\,(BE+CD)\quad\quad\dots(1)}$

Here,

• $\sf{OA=BE=2\ N}$

• $\sf{AB=2\ m}$

• $\sf{DE=1\ m}$

• $\sf{CD=3\ N}$

Hence (1) becomes,

$\longrightarrow\sf{K=2\times2+\dfrac{1}{2}\times1\,(2+3)}$

$\longrightarrow\sf{K=4+\dfrac{5}{2}}$

$\longrightarrow\sf{\underline{\underline{K=6.5\ J}}}$

Answer 2

area under the f-s graph will give the work or change in work

here

area of f-s graph=area of the square from the front,+ area of trapezium formed

=2×2+1/2 ×(2+3) ...( area of trapezium is 1/2(sum of //side)×H)

=4+2.5

=6.5

change in work=change in KE

so

since intial KE zero,so change in KE is final KE

KE=6.5

hope it helps