Question

A constant force is applied on a $$10\ kg$$ mass at rest. The ratio of works done in $$1st, 2nd$$ and $$3rd$$ seconds is

A
1:2:3
B
1:3:5
C
1:4:9
D
1:1:1

Solution

The correct option is B $$1 : 3 : 5$$$$\textbf{Step 1: Calculation of acceleration}$$ Applying Newton's Law on $$10\ kg$$ mass                        $$\sum F = ma$$        and     $$\sum F = F$$                  $$\Rightarrow\ a = \dfrac{F}{m}$$ As force is constant, hence acceleration will be constant during the whole motion. $$\textbf{Step 2: Distance travelled in}\ 1^{st} , 2^{nd}\ \textbf{and}\ 3^{rd}\ \textbf{second}.$$For constant acceleration, Distance travelled in $$n^{th}$$ second is given by:                    $$S_{n} = u + \dfrac{a}{2} [2n - 1]$$,    Here $$u = 0 \Rightarrow S_{n} = \dfrac{a}{2} [2n - 1]$$ In $$1^{st}$$ second, $$n = 1 \quad \Rightarrow \quad S_{1} = \dfrac{a}{2} [2 (1) - 1] = \dfrac{a}{2}$$ In $$2^{nd}$$ second, $$n = 2 \quad \Rightarrow \quad S_{2} = \dfrac{a}{2} [2 (2) - 1] = \dfrac{3a}{2}$$ In $$3^{rd}$$ second, $$n = 3 \quad \Rightarrow \quad S_{3} = \dfrac{a}{2} [2 (3) - 1] = \dfrac{5a}{2}$$                                         $$\Rightarrow \quad S_{1} : S_{2} : S_{3} = 1 : 3 : 5$$ Hence Option $$(B)$$ is correct.Physics

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