Question

20.
The work done in time 't' on a body of mass 'm' which is accelerated from rest to a speed 'v' in time t,'
as a function of time 't' is given by

​

Answer 1

Answer:

Make sure Velocity V in time t1

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Thanks for Reading

Answer 2

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

The work done in time 't' on a body of mass 'm' which is accelerated from rest to a speed 'v' in time 't₁' as function of time 't' is given by?

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

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

• Initial velocity (u) = 0 m/s.
• Final velocity (v) = v m/s .

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

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The Velocity "v" is in time t₁ seconds.

Applying First kinematics equation,

$\large{\boxed{\tt v = u + at}}$

Here,

• u = 0 m/s.
• t = t₁ seconds
• v = v m/s.

Substituting the values,

$\large{\tt \leadsto v = 0 + a \times t_1}$

$\large{\tt \leadsto v = a \times t_1}$

$\large{ \leadsto {\underline{\underline{\tt a = \dfrac{v}{t_1}}}} \: \tt ------(1)}$

Now, Applying Newton's second law of motion,

$\large{\boxed{\tt F = ma}}$

Substituting the values of Acceleration from Equation (1).

$\large{\tt \leadsto F = m \times \dfrac{v}{t_1}}$

$\large{\leadsto {\underline{\underline{\tt F = m.\dfrac{v}{t_1}}}} \: \tt ------(2)}$

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Now, Finding the Displacement in time "t" seconds.

From Second kinematic equation,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values of Acceleration from equation (1).

$\large{\tt \leadsto S = 0 \times t + \dfrac{1}{2} \times \dfrac{v}{t_1} \times t^2}$

$\large{\tt \leadsto S = 0 + \dfrac{v}{2t_1} \times t^2}$

$\large{\leadsto {\underline{\underline{\tt S = \dfrac{v}{2t_1}.t^2}}} \: \tt ------(3)}$

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Now,Applying Work done formula,

$\large{\boxed{\tt W = F.s}}$

Substituting the values From equation 2 & 3.

$\large{\tt \leadsto W = m.\dfrac{v}{t_1} \times \dfrac{v}{2t_1}.t^2}$

$\large{\tt \leadsto W = m.\dfrac{v^2}{t_1} \times \dfrac{1}{2t_1}.t^2}$

$\huge{\boxed{\boxed{\tt W = \dfrac{mv^2}{2(t_1)^2}t^2}}}$

Hence derived!

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