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