Question

A block of mass $10\;kg$ moves in X-direction with a constant speed of $10ms^{-1}$, is subjected to a retarding force $F=-0.1xJ/m$ during its travel from $x=-20m$ to $30\;m$. Its final KE will be:

A) 275 J  

B) 250 J 

C) 475 J 

D) 450 J

Answer 1

Hint: We know that the work is defined as the product of force and displacement. Here we have a force which displaces the particle in the x direction. Hence to calculate the work, we must calculate the displacement of the particle and then use scalar multiplication to find the work done.

Formula used:

$W=F\cdot d$

Complete step-by-step answer:

We know that work done is the scalar product of force $F$ and the displacement $d$. And is given as $W=F\cdot d=Fdcos\theta$ where $\theta$ is the angle between the force $F$ and the displacement $d$. Since it is given that the body moves in the x-direction, we can say that $\theta=0$ and thus $cos(0)=1$

Then we can say that the work done by the spring is nothing but the change in the kinetic energy of the spring. That is to say that, $W=\Delta KE$

Then for a small displacement $d\;x$ the small work done $d\;W$ is given as $dW=\int F.dx$

Here, it is given that the $F=0.1 xJ/m$ and the displacement is along the x-axis from $-20\;m$ to $30\;m$. Also, given that $m=10kg$ and moves with initial velocity $u=10m/s$

Then, we have $\int F.dx=\dfrac{1}{2}m(v^{2}-u^{2})$

$\implies \int_{-20}^{30}- 0.1x dx=KF_{f}-\dfrac{1}{2}10(10)^{2}$

$\implies \dfrac{30^{2}-(-20)^{2}}{2}(-0.1) =KF_{f}-\dfrac{1}{2}10(10)^{2}$

$\implies 500\times\dfrac{-0.1}{2}+\dfrac{1000}{2}=KF_{f}$

$\implies KE_{f}=475$

Hence the correct answer is option (C): 475 J

Note: The scalar multiplication can be used to multiply a scalar and a vector or two vectors. If there are two vectors $\vec A$ and $\vec B$, then the scalar product is defined as $\vec A \cdot \vec B=|\vec A||\vec B|cos\theta$, where $\theta$ is the angle between the $\vec A$ and $\vec B$. Here we are integrating to get the work done, the student must know some basic integration to solve this sum. The scalar multiplication is also known as the dot product. Also, the resultant of scalar multiplication is always a scalar.