Question

A box of mass 72 kg is at rest on a horizontal frictionless surface. A constant horizontal force F then acts on the box and accelerates it to the right. It is observed that it takes the box 3.4 seconds to travel 13 meters. What is the magnitude of the force

Answer 1

Answer:

$162\ N$

Explanation:

$We\ are\ given\ that,\\Mass\ of\ the\ box=72\ kg\\Distance/Displacement\ travelled\ by\ the\ box\ during\ the\ application\ of\\ Force=13\ m\\Time\ taken\ for\ it\ to\ displace=3.4 \seconds\\Now,\\As\ we\ know\ that,\\Force=Mass*Acceleration\\In\ order\ to\ know\ force\ we\ need\ to\ know\ the\ acceleration.\ Lets\ find\\ that\ out.\\We\ would\ use\ Newton's\ Third\ Equation\ Of\ Motion\ and\ Newton's\\ First\ Equation\ Of\ Motion:\\2as=v^2-u^2\\v=u+at$

$Hence,\\First\ lets\ consider\ Newtons\ First\ Equation\ Of\ Motion:\\v=u+at\\v-u=at\\Hence\ now\ lets\ move\ on\ to\ Newton's\ Third\ Law\ Of\ Motion:\\2as=v^2-u^2\\2as=(v+u)(v-u)\\Substituting\ (v-u)=at,\\2as=at(v+u) \\Hence,\ as\ the\ body\ moves\ from\ rest,\ u=0\\So,\\2as=at*v\\Cancelling\ a\ at\ both\ the\ sides\ we\ get,\\2s=vt\\Hence,\\Lets\ plug\ in\ the\ values\ of\ Displacement\ and\ Time,\ Shall\ we?\\Hence,\\2*13=3.4v\\26=3.4v\\\frac{26}{3.4}=v\\v \approx 7.647\ m/s$

$Now,\\As\ we\ know\ that,\\a=\frac{v-u}{t} [Equation\ for\ acceleration]\\Hence,\\a=\frac{v}{t} [As\ u=0]\\Hence,\\Acceleration=\frac{7.647}{3.4} \\Acceleration \approx 2.25\ m/s^2$

$Hence,\\Now,\\As\ by\ using\ expression\ for\ Force,\\Force= Mass*Acceleration\\Here,\\Force\ exerted\ on\ the\ box=72*2.25= 162 N$