Question

A varying force acting on a 4-kg object causes it it have displacement given by $x=2t-3t^2+t^3$ (where x is in meters and t is in seconds). The object starts at rest. Find the work this force does on the object in the first 3s of motion

• A. 234 J
• B. 266 J
• C. 422 J
• D. 655 J

Answer 1

The correct option is A 234 J

$x=2t-3t^2+t^{3}dx=(2-6t+3t^2)dt\dots \dots \left(1\right)v=\frac{dx}{dt}=(2-6t+3t^2)Acceleration,a=\frac{dv}{dt}=-6+6t=6(t-1)F=ma=4\times 6(t-1)=24(t-1)$
Work done for small displacement dx is
$dW=Fdx=F=(2-6t+3t^2)dt\therefore Workdonein3sisW=24{\int }_0^3(t-1)(2-6t+3t^2)dt=24{\int }_0^3(3t^3-9t^2+8t-2)dt=24{[\frac{3t^4}{4}-3t^3+4t^2-2t]}_0^3=24[\frac{243}{4}-81+36-6]orW=39\times 6=234J$