Question

work done by a body is function at time and is given as w=3t^3+4t^2+8 calculate the power of the body please write steps and explain

Answer 1

Given:

Work done by a body,W= 3t³+4t²+8

To Find:

Power applied by body

Solution:

We know that,

• Power P is given by

$\pink{\underline{\boxed{\bf{P=\frac{dW}{dt}}}}}$

where,

W is work done

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Let the power applied by body be P

So,

$\longrightarrow\rm{P=\dfrac{dW}{dt}}$

$\longrightarrow\rm{P=\dfrac{d(3t^{3}+4t^{2}+8)}{dt}}$

$\longrightarrow\rm{P=3(3t^{2})+4(2t)+0}$

$\longrightarrow\rm\green{P=9t^{2}+8t}$

Hence, the power applied by given body is 9t²+8t where t is time taken by body.

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Additional Information

Power

Power can be defined as the rate or capacity of doing work.

It is a scalar quantity.

It's S.I. unit is watt(W)

Work

Work can be defined as dot product of force applied and displacement of body.

It is a scalar quantity.

It's S.I. unit is joule(J)

Answer 2

Answer

w = ${3t}^{3} + {4t}^{2} + 8$

$p = \frac{dw}{dt}$

$\longrightarrow$Differentiation of equation using $\frac{d( {x}^{n} )}{dx} = nx ^{ - 1}$

$\longrightarrow$${9t}^{2} + 8t$

The power of a body at any time t is given by the substituting the value of t in the equation ${9t}^{2} + 8t.$

ADDITIONAL INFORMATION :-

$\longrightarrow$$\frac{d(c)}{dx} = 0$

$\longrightarrow$$\frac{d(sinx)}{dx} = cosx$

$\longrightarrow$$\frac{d(cosx)}{dx} = - sinx$

$\longrightarrow$$\frac{d(logx)}{dx} = \frac{1}{x}$

$\longrightarrow$$\frac{d( {e}^{x}) }{dx} = {e}^{x}$

Thanks