Question

A particle moves along the x-axis from x = 1 to x = 3 m under the influence of a force F = (3 x² - 2x + 5) N. Find the work done in this
process.​

Answer 1

Work done = 28 J

Given

A particle moves along the x-axis from x = 1 to x = 3 m under the influence of a force F = (3 x² - 2x + 5) N

To Find

Work done

Concept Used

Work done is defined as product of force and displacement

$\bf \pink{\bigstar\ \; W=\int\limits^a_b Fdx}$

SI unit of Work is Joule

SI unit of Force is Newton

SI unit of displacement is Meter

Solution

$\bf \displaystyle W=\int^3_{1}(3x^2-2x+5)dx\\\\\to \rm \displaystyle W=\bigg[x^3-x^2+5x\ \; \bigg]_1^{3}\\\\\to \rm \displaystyle W=[3^3-3^2+5(3)]-[1^3-1^2+5(1)]\\\\\to \rm \displaystyle W=[27-9+15]-[1-1+5]\\\\\to \rm \displaystyle W=33-5\\\\\to \bf \green{\displaystyle W=28\ J\ \; \bigstar}$

So , Work done = 28 J

Answer 2

Explanation:

Work done = 28 J

Given

A particle moves along the x-axis from x = 1 to x = 3 m under the influence of a force F = (3 x² - 2x + 5) N

To Find

Work done

Concept Used

Work done is defined as product of force and displacement

\bf \pink{\bigstar\ \; W=\int\limits^a_b Fdx}★ W=

b

∫

a

Fdx

SI unit of Work is Joule

SI unit of Force is Newton

SI unit of displacement is Meter

Solution

\begin{gathered}\bf \displaystyle W=\int^3_{1}(3x^2-2x+5)dx\\\\\to \rm \displaystyle W=\bigg[x^3-x^2+5x\ \; \bigg]_1^{3}\\\\\to \rm \displaystyle W=[3^3-3^2+5(3)]-[1^3-1^2+5(1)]\\\\\to \rm \displaystyle W=[27-9+15]-[1-1+5]\\\\\to \rm \displaystyle W=33-5\\\\\to \bf \green{\displaystyle W=28\ J\ \; \bigstar}\end{gathered}

W=∫

1

3

(3x

2

−2x+5)dx

→W=[x

3

−x

2

+5x ]

1

3

→W=[3

3

−3

2

+5(3)]−[1

3

−1

2

+5(1)]

→W=[27−9+15]−[1−1+5]

→W=33−5

→W=28 J ★

So , Work done = 28 J