Question

A force F=a+b x 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:

• The Work done (W) on the particle is a d + ( b d² / 2 )

Given:

1. The given Relation is F = a + b x
2. Displacements given x₁ = 0 and x₂ = d

Explanation:

$\rule{300}{1.5}$

From the relation we know,

$\large \bigstar\; \boxed{\tt W = \displaystyle\int\limits^{x_2}_{x_1} \tt F.dx}$

$\frak{Here}\begin{cases}\text{W Denotes Work Done}\\\text{F Denotes the variable Force}\end{cases}$

Now,

$\large \boxed{\tt W = \displaystyle\int\limits^{x_2}_{x_1} \tt F.dx}$

Substituting the values,

$\displaystyle \dashrightarrow\tt W=\displaystyle\int\limits^{x_2}_{x_1} \tt (a+bx).dx\\\\\\\dashrightarrow\tt W=\displaystyle\int\limits^{d}_{0} \tt (a+bx).dx\\\\\\\dashrightarrow\tt W=\displaystyle\int\limits^{d}_{0} \tt a.dx+\displaystyle\int\limits^{d}_{0} \tt bx.dx\\\\\\\dashrightarrow\tt W=\Bigg[a\;x\Bigg]^d_0+\Bigg[\dfrac{b\;x^2}{2}\Bigg]^d_0\\\\\\\dashrightarrow\tt W=\Bigg[a\;(d-0)\Bigg]+\Bigg[\dfrac{b\;(d^2-0^2)}{2}\Bigg]\\\\\\\dashrightarrow\tt W= a\times d+\dfrac{b\times d^2}{2}$

$\displaystyle\dashrightarrow \large{\underline{\boxed{\red{\tt W=a\;d+\dfrac{b\;d^2}{2}}}}}$

∴ The Work done (W) on the particle is a d + ( b d² / 2 ).

$\rule{300}{1.5}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

As we know that :

$\huge{\boxed{\boxed{\sf{W \: = \: \int F. dx}}}}$

Where, F = a + dx

Substitute value of F,

$\implies {\sf{W \: = \: \int (a \: + \: bx). dx}} \\ \\ \implies {\sf{W \: = \: \int a dx \: + \: \int d bx^2}} \\ \\ \implies \displaystyle {\sf{W \: = \: \bigg[ ax \bigg]_0 ^d\: + \: \bigg[ b \dfrac{x^2}{2}\bigg]_0 ^d}} \\ \\ \implies {\sf{W \: = \: \bigg[ a(d \: - \: 0) \bigg] \: + \: \bigg[ \dfrac{b(d^2 \: - \:0)}{2}\bigg]}} \\ \\ \implies {\sf{W \: = \: ad \: + \: \dfrac{bd^2}{2}}} \\ \\ \implies {\sf{W \: = \: \dfrac{2ad \: + \: bd^2}{2}}} \\ \\ {\LARGE{\boxed{\sf{W \: = \: \dfrac{2ad \: + \: bd^2}{2}}}}}$