Question

8.
A force F = kx² acts on a particle at an angle of 60° with
the x-axis. The work done in displacing the particle from x1,
to x2, will be -
5-0 (1) ky ?
(3) * cx3 = x})
(2)(x3 x})
(4) 5(x3 -x})
3
NW​

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A force f = kx² acts on a particle at an angle of 60° with the x-axis. The work done in displacing the particle from x₁ to x₂ will be?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Angle Made by the Force Vector is 60°.
• Given Force relation = kx².

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

As stated in the Question Force is making an angle 60° with the x - axis ,

Therefore we need to take the force component along x - axis and That will be f cosθ.

Now,

⇒ F = f cos60°

⇒ F = kx² × ½ [cos60° = ½]

⇒ F = ½ kx² N.

(This Force will cause the body to move in x - axis or x - direction)

$\rule{300}{1.5}$

$\rule{300}{1.5}$

As the Displacement is Very small,

Therefore,

$\large{\boxed{\tt W = \displaystyle\int \tt F.dx}}$

Substituting the values,

$\large{\tt W = \displaystyle\int \tt \bigg(\dfrac{1}{2} kx^2 \bigg) .dx}$

Applying the limits I.e x₁ and x₂.

$\large{\tt W = \displaystyle \int_{\tt x_{1}}^{\tt x_{2}} \tt \bigg(\dfrac{1}{2} kx^2 \bigg) .dx}$

Integrating,

( ½ k is a Constant and cannot be integrated)

$\large{\tt W = \dfrac{1}{2}.k \Bigg[\dfrac{x^3}{3} \Bigg]_{x_{1}}^{x_{2}}}$

The integration Formula used is:-

$\large{\boxed{ \displaystyle\int \tt x^n dx = \dfrac{x^{n + 1}}{n + 1} \: \rightarrow n \neq - 1}}$

Simplifying,

$\large{\tt W = \dfrac{1}{2}.k \Bigg[\dfrac{x^3}{3} \Bigg]_{x_{1}}^{x_{2}}}$

$\large{\tt W = \dfrac{1}{2}.k \Bigg[\dfrac{x_2^3}{3} - \dfrac{x_1^3}{3} \Bigg]}$

$\large{\tt W = \dfrac{1}{2}.k \times \dfrac{1}{3}[x_2^3 - x_1^3]}$

$\huge{\boxed{\boxed{\tt W = \dfrac{1}{6}.k [x_2^3 - x_1^3]}}}$

Therefore, work done is W = ⅙. k[x₂³ - x₁³]

$\rule{300}{1.5}$