Question

17. A particle of mass 2 kg travels along a straight line
with velocity v=a(root)x where a is a constant. The
work done by net force during the displacement of
particle from x = 0 to x = 4 m is
(1) a^2
(2) 2a^2
(3) 42²
(4) √2a²​

Answer 1

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

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

$\large{\bold{v = a \sqrt{x}}}$

m = 2 Kg.

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

As you know,

$\large{\bold{a = \frac{dv}{dt}}}$

As velocity is given in Distance we need to convert the formula so that x comes in denominator.

Now,

$\large{ \implies a = \frac{dv}{dt} \times \frac{dx}{dx}}$

$\large{ \implies a = \frac{dx}{dt} \times \frac{dv}{dx}}$

$\large{\bold{a = v. \frac{dv}{dx}}}$

Now, Substituting the values.

$\large{ \implies a = v. \frac{d(a \sqrt{x})}{dx}}$

$\large{ \implies a = v. \frac{1}{2}a.x^{ (\frac{- 1}{2})}}$

It becomes,

$\large{ \implies a = v. \frac{a}{2 \sqrt{x}}}$

Now, substituting the values of velocity.

$\large{ \implies a = \frac{a}{2 \sqrt{x}} \times a \sqrt{x}}$

$\large{ \implies a = \frac{a}{2 \cancel{\sqrt{x}}} \times a \cancel{ \sqrt{x}}}$

$\large{\boxed{a = \frac{a^2}{2}}}$

Applying Newton's second law.

$\large{\bold{F = ma}}$

Substituting the values.

$\large{ \implies F = 2 \times \frac{a^2}{2}}$

$\large{ \implies F = \cancel{2} \times \frac{a^2}{ \cancel{2}}}$

$\large{\boxed{F = a^2}}$

As we know,

$\large{\bold{W = \displaystyle\int F.ds}}$

$\large{ \implies W = \displaystyle\int a^2.ds}$

Applying limits.

$\large{ \implies W = \displaystyle\int^4_0 a^2.ds}$

As a is constant it will come out.

$\large{ \implies W = a^2 \displaystyle\int^4_0 ds}$

$\large{ \implies W = a^2 \left|x \right|^4_0}$

Substituting,

$\large{ \implies W = a^2 \times (4 - 0)}$

$\huge{\boxed{\boxed{W = 4a^2 \: J}}}$

So, the work done by the force is 4a².

Answer 2

$\huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

• Mass of the particle,m = 2 Kg

• Velocity,v = a√x

Let ā be the acceleration of the particle

We know that,

$\boxed{ \boxed{\huge{ \sf{a = v. \frac{dv}{dx} }}}}$

Differentiating v w.r.t to t,we get:

$\sf{ \bar{a} = \frac{d(a \sqrt{x}) }{dx}.a \sqrt{x} } \\ \\ \implies \: \sf{ \bar{a} = a \cancel{\sqrt{x}}. \frac{a}{2 \cancel{\sqrt{x} }} } \\ \\ \implies \: \huge{\sf{ \bar{a} = \frac{a {}^{2} }{2} }}$

From Newton's Second Law of Motion,

$\sf{f \: = m \bar{a}} \\ \\ \rightarrow \: \sf{f = 2. \frac{a {}^{2} }{2} } \\ \\ \rightarrow \: \underline{ \boxed{ \sf{f = a {}^{2} }}}$

We know that,

W = Force × Displacement

W = F.s

Refer to the attachment