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  • The force on a particle of mass is <img alt="\mathrm{(10 \hat{i

The force on a particle of mass \mathrm{10 \mathrm{~kg}} is \mathrm{(10 \hat{i}+5 \hat{j}) \mathrm{N}}. If it starts from rest, the position at time \mathrm{t=5\, \, \mathrm{sec}} __________

Option: 1

\mathrm{100 \hat{\imath}+600 \hat{\jmath}}\, \,


Option: 2

\mathrm{12500 \, \hat{i}+6250\, \hat{j}}


Option: 3

\mathrm{12000 \, \hat{i}+6200 \, \hat{j}}


Option: 4

\mathrm{1000 \, \hat{i}+5000 \, \hat{j}}


Answers (1)

best_answer

We have, \mathrm{F_x=10 \mathrm{~N}} 

\mathrm{ a_x=\frac{F_x}{m}=\frac{10 \mathrm{~N}}{0.01 \mathrm{~kg}}=1000 \mathrm{~m} / \mathrm{s}^2 }

As this is a case of constant acceleration in the x-direction.

\mathrm{ \begin{aligned} & x=u_x t+\frac{1}{2} a_x t^2=\frac{1}{2} \times 1000 \mathrm{~m} / \mathrm{s}^2 \times(5 \mathrm{sec})^2 \\ & x=12500 \mathrm{~m} \end{aligned} }

Similarly,  \mathrm{a_y=\frac{f_y}{\mathrm{~m}}=\frac{5 \mathrm{~N}}{0.01 \mathrm{~kg}}=500 \mathrm{~m} / \mathrm{s}^2}

               \mathrm{y=u_y t+\frac{1}{2} a_y t^2=6250 \mathrm{~m}}

Thus, the position of the particle at \mathrm{t=5\, \mathrm{sec}} is -

\mathrm{\vec{r}=(12500 \hat{i}+6250 \hat{j}) \mathrm{m} \text { Ans. }}

Posted by

Ajit Kumar Dubey

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