Question

A force of 30 N acts on a body of mass 2.0 kg starting
from rest upto a distance of 3.0 m. Then the force
reduces to 15 N and acts in the same direction upto
2.0 m. Calculate the final kinetic energy of the body.​

Answer 1

$\huge\mathtt{★\:Solution\: :-}$

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$\tt{For \:the\: total \:3m\: distance: \:F\: =\: ma}$

↦$\tt{a\: =}$$\tt\frac{F}{m}$$\tt{=\: 30/2 \: =\:15m\: /}$$\tt{s}^{2}$

↦$\tt{v}^{2}\: =$$\tt{u}^{2}\: +\:2\:as$

↦$\tt{=\:0\:+\:2\:×\:15\:×\:3}$

↦$\tt{v}^{2}\: =\: 90$

↦$\tt{v\: =}$ $\tt{√90m/s}$

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$\tt{For \:the\: next \:2m \:distance,}$

↦$\tt{a′ \:=}$$\tt\frac{F′}{m}\: =$$\tt\frac{15}{2}\:=$$\tt{7.5\:m\: /}$$\tt{s}^{2}$

↦$\tt{v′}^{2}\: =$$\tt{v}^{2}$$\tt{+\:2\:as}$

↦$\tt{90\:+\:2\:×\:7.5\:×\:2\:=\:90\:+}$$\tt{30\:=\:120}$

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$\tt{Final \:kinetic\: energy\: =}$$\tt\frac{1}{2}\:m$$\tt{v}^{2}$

↦$\tt\frac{1}{2}\:×\:2\:×\:120$$\tt{=\:120J}$

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$\tt\blue{Answer\:⇝}$$\tt\pink{120\:J}$