Question

A force F = a + bx acts on a particle in the x-direction, where a and b are constants. Find the work done by this force during a displacement from x = 0 to x = d.

Answer 1

Answer:

A force F = a + bx acts on a particle in the x-direction, where a and b are constants. Find the work done by this force during a displacement from x = 0 to x = d.

Answer 2

The work done by this force during a displacement from x = 0 to x = d is $\mathrm{W}=\int \mathrm{d} 0 \mathrm{Fdx}=\mathrm{axd}+\mathrm{b} \times \mathrm{d} 22 \mathrm{J}$

Explanation:

Step 1:

On the particle, the acting force is $F=a+b * x$.

Here, the constants are a and b. So, the force performs the work during the displacement, x = 0 and x = d is shown as

$W=\int_{0}^{d} F d x=\int_{0}^{d}(a+b x) d x=a x+\left(b x^{2} / 2\right)=[a+1 / 2 b d] d$

$W=\vec{F} \cdot \vec{r}=F r \cos \theta$

Step 2:

Here, the angle between the direction of displacement and the force is denoted as θ.

In the moving particle, the work done is denoted as $\mathrm{dW}=\mathrm{Fdx}=\mathrm{Fdx}=\left(\mathrm{b}^{*} \mathrm{x}+\mathrm{a}\right) \mathrm{dx}$

In moving the particle, the work done from x = d m is:  

$\mathrm{W}=\int \mathrm{d} 0 \mathrm{Fdx}=\mathrm{axd}+\mathrm{b} \times \mathrm{d} 22 \mathrm{J}$