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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A force of $(6 \hat i+8 \hat j) N $ acted on a body of mass 10 kg.The displacement after 10 sec. if it starts from rest will be

a) 50 m along $ tan ^{-1} 4/3 $ with x axis b) 70 m along $ tan ^{-1} 3/4 $ with x axis c) 10 m along $ tan ^{-1} 4/3 $ with x axis d) None
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1 Answer

Answer: 50m along $\tan^{-1} \large\frac{4}{3}$ with x-axis
Acceleration $a=\large\frac{\hat F}{m}=\large\frac{6 \hat i+8 \hat j}{10}$
Since we know the acceleration, we can use the Kinematics equation $ \overrightarrow s=\overrightarrow u t+\large\frac{1}{2} $$ \overrightarrow a t^2$ to get the displacement.
$\quad=0+\large\frac{1}{2}\bigg(\frac{6 \hat i+8 \hat j}{10}\bigg)$$10 ^2 $$=30 \hat i+40 \hat j$
Magnitude $= \sqrt {30 ^2 +40 ^2}= 50\;m$
The direction is along an angle of $\tan ^{-1}\large\frac{40}{30}$$=\tan ^{-1} \;\large\frac{4}{3}$ with x-axis
answered Jul 4, 2013 by meena.p
edited Aug 20, 2014 by balaji.thirumalai
 

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