Question

A force $F=(3x+4)$ displaces an object of mass $10kg$ from $x=0$ to $x=4m$ if the speed of the object at $x=0$ is 2m/sec then its speed at x=4m will be

• A. $\sqrt{3}m/s$
• B. $2\sqrt{3}m/s$
• C. $3m/s$
• D. $4m/s$

Answer 1

The correct option is B

$2\sqrt{3}m/s$

Step1: Given data

A force, $F\left(x\right)=(3x+4)$displaces an object of mass, $m=10kg$ from x=0 i.e. $x_i=0$ to x=4m i.e. $x_f=4m$.

If the speed of the object at $x=0$ is $2m/s$ i.e. $v_i=2m/s$

Step2: Formula used

Work done, $W={\int }_{x_i}^{x_f}F\left(x\right)dx$

Here, $F\left(x\right)$ is the force which is the function of displacement $x$ and $dx$ is the small displacement, $x_i$ and $x_f$ is the initial and final position of the object.

According to the work-energy theorem, the work done is the change in kinetic energy.

Work-energy theorem, $W=∆K=\frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2$

where, $m$ is the mass, $v_f$ is the final velocity and $v_i$ is the initial velocity

Step3: Calculating work done

$W={\int }_{x_i}^{x_f}F\left(x\right)dx$

Substituting $F\left(x\right)=(3x+4)$ in the above equation, we get,

$$\begin{gathered} W={\int }_0^4(3x+4)dx \\ W={\left(\frac{3x^2}{2}+4x\right)}_0^4 \\ W=\left(\frac{3{\left(4\right)}^2}{2}+4\left(4\right)\right)-(0+0) \\ W=24+16 \\ W=40J \end{gathered}$$

Step4: Calculating final speed

$W=∆K=\frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2$

Substituting $W=40J$ , $m=10kg$ and $v_i=2m/s$ in the above equation, we get,

$$\begin{gathered} 40=\frac{1}{2}\left(10\right){v_f}^2-\frac{1}{2}\left(10\right){\left(2\right)}^2 \\ 40+20=5{v_f}^2 \\ {v_f}^2=12 \\ {v}_f=2\sqrt{3}m/s \end{gathered}$$

Therefore, the final velocity of the object will be $2\sqrt{3}m/s$.

Hence, option B is the correct option.