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Q. An object of mass $5 kg$ is acted upon by a force that varies with the position of the object as shown in the figure. If the object starts out from rest at the point $x=0$, then its speed at $x=25\, m$ is
image

Laws of Motion

Solution:

Till $x=25 m , F=10 N$
$\Rightarrow g_{30}=\frac{F}{m r }=\frac{10 N }{5 kg }=2 \,ms ^{-2}$
Using $v^{2}-u^{2}=2 a s$
$v^{2}-0^{2}=2(2)(25), v^{2}=100$ or $v=10 \,ms ^{-1}$