Question

A force F = 20+10y acts on a particle in y direction where F is in newton and y in meter. work done by this force to move the particle from y=0 to y=1 m is? ​

Answer 1

workdone is the dot product of force and displacement. and when force is function of displacement then, workdone is found by using the concept of integration.

workdone = $\int\limits^1_0{F(y)}\,dy$

= $\int\limits^1_0{(20+10y)}\,dy$

= $20\int\limits^1_0{dy}+10\int\limits^1_0{y}\,dy$

= 20[1 - 0] + 10[1²/2 - 0²/2 ]

= 20 + 5

= 25 Joule

hence, workdone by force to move particle from 0 to 1 is 25 Joule.

Answer 2

Answer:

25 Joules.

Explanation:

Given:

Force on the particle, $F = 20+10y$.

The work done on an object in displacing it to a small displacement $\vec d$ is given by

$W = \vec F \cdot \vec d$

In the given question, the force is given along the y direction, therefore,

$\vec F \cdot \vec d = F\ y$

But the given force is varying with distance,

The small work done in displacing the particle to a small distance $dy$ is given by

$dW = F\ dy$.

The total work done in moving the particle from $y=0\ m$ to $y=1\ m$ is given by integrating the above expression with respect to $y$, within the limits $y=0\ m$ to $y=1\ m$,

$W = \int dW \\=\int^{y=1}_{y=0} F\ dy\\=\int^{y=1}_{y=0}(20+10y)\ dy\\=\left( 20y+ 10\dfrac{y^2}{2} \right )^{y=1}_{y=0}\\=\left( (20\cdot 1-20\cdot 0)+ 10\left ( \dfrac{1^2}{2}-\dfrac{0^2}{2}\right ) \right )\\=\left( 20+ \dfrac{10}{2} \right )\\=25\ J.$