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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 140 N. How far does the crate travel in 10.05 s?

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Answer: 216 m
Let us set up the problem first:
Given $F, m$ we first need to calculate $a_x$.
From Newton's 2nd law, $F = 140 N = ma_x = 32.5 \times a_x \rightarrow a_x = \large\frac{140}{32.5}$$ = 4.31\;m/s^2$
Given that we have $a_x$, we can now calculate the distance traveled from the kinematics equation: $x = x_0 + V_0t + \large\frac{1}{2}$$a_xt^2$
$\Rightarrow x = 0+ 0 \times 10.05 + \large\frac{1}{2}$$ 4.31 \times 10.05^2 = 216\;m$
answered Aug 19, 2014 by balaji.thirumalai

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