A force F1 acting on a 2Kg body produces an acceleration of 2-51exm- -1exs2- Another force F2 acting on a 5Kg body produces an acceleration of 21exm- -1exs2- The ratio 1exF2- -1exF1-is-

Step1 Given quantities:m_1=2Kg0ex0exa_1=2.51exm/ 1exs^2.0ex0exm_2=5Kg0ex0exa_2=21exm/ 1exs^2.Step2 Formula Used:F_2/F_1=m_2×a_2/m_1×a_10ex0exF_1→ Force on fir ...

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Byju's Answer

Standard VIII

Physics

Free Body Diagram

A force F1 ac...

Question

A force $F_{1}$ acting on a $2 K g$ body produces an acceleration of $2.5 \frac{m}{s^{2}}$ . Another force $F_{2}$ acting on a $5 K g$ body produces an acceleration of $2 \frac{m}{s^{2}}$ . The ratio $\frac{F_{2}}{F_{1}}$ is:

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Solution

The correct option is A
2

Step1 Given quantities:
$m_{1} = 2 K g a_{1} = 2.5 \frac{m}{s^{2}} m_{2} = 5 K g a_{2} = 2 \frac{m}{s^{2}}$
Step2 Formula Used:
$\frac{F_{2}}{F_{1}} = \frac{m_{2} \times a_{2}}{m_{1} \times a_{1}} F_{1} \to F o r c e o n f i r s t b o d y F_{2} \to F o r c e o n s e c o n d b o d y m_{2} \to m a s s o f s e c o n d b o d y a_{2} \to a c c e l e r a t i o n o n s e c o n d b o d y m_{1} \to m a s s o f f i r s t b o d y a_{1} \to a c c e l e r a t i o n o n f i r s t b o d y$
Step3 The calculation for the ratio:
$\frac{F_{2}}{F_{1}} = \frac{5 \times 2}{2 \times 2.5} \frac{F_{2}}{F_{1}} = \frac{10}{5} \frac{F_{2}}{F_{1}} = 2$
Hence, option (A) is correct the correct answer.

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A force F1 acting on a 2Kg body produces an acceleration of 2.5ms2. Another force F2 acting on a 5Kg body produces an acceleration of 2ms2. The ratio F2F1is:

A force $F_{1}$ acting on a $2 K g$ body produces an acceleration of $2.5 \frac{m}{s^{2}}$ . Another force $F_{2}$ acting on a $5 K g$ body produces an acceleration of $2 \frac{m}{s^{2}}$ . The ratio $\frac{F_{2}}{F_{1}}$ is: