Question

A particular force on an object applies such that its strength can be expressed as a function of the object's displacement as $F\left(x\right)=6x^2-2x-5$ where $F$ is in Newtons and $x$ is in meters.
The object has $32J$ of kinetic energy when it is at the origin. From there, it is displaced $4m$.
What is its final kinetic energy?

• A. $32J$
• B. $60J$
• C. $92J$
• D. $124J$
• E. $144J$

Answer 1

Answer: (D)

$124J$

The force acting on the object $F\left(x\right)=6x^2-2x-5$ N

Work done by force $W=\int F\left(x\right)dx$

$\therefore$ Work in moving from $x=0$ to $x=4$ m, $W={\int }_o^4(6x^2-2x-5)dx$

OR $W=(2x^3-x^2-5x){|}_0^4=\left[2\right(4{)}^3-(4{)}^2-5(4\left)\right]-0=92$ J

Initial kinetic energy of the object $K.E_i=32$ J

Work-energy theorem : $W=K.E_f-K.E_i$

$\therefore$ $92=K.E_f-32$ $⟹K.E_f=124$ J