Question

Give that the displacement of the body in meter is a function of time as follows x=2t power to 4 +5 the mass of the body is 2kg. What is the increase in it's kinetic energy one second after the start of motion?

Answer 1

Answer:

Explanation:

v= dx/dt

v= d(2t⁴ + 5)/dt

v= 8t³  ,

at t =1  , your v will be = 8 m/s.

increase in K.E = final k.e -  intial k.e                     (initial K.E = 0 )

increase in K.E =   1/2*m*v²  - 0              

                         = 1/2*2*(8)²

                         =  64 J

Answer 2

The increase in it's kinetic energy one second after the start of motion is 64 J.

Explanation:

Given that,

Mass of the body = 2kg

The displacement of the body  is

$x=2t^{4}+5$

On differentiating w.r.to x

$v=\dfrac{dx}{dt}$

$v=\dfrac{d}{dt}(2t^4+5)$

$v=8t$

We need to calculate the increase in it's kinetic energy one second after the start of motion

Using formula of kinetic energy

$K.E=\dfrac{1}{2}m(v_{f}^2-v_{i}^2)$

Put the value into the formula

$K.E=\dfrac{1}{2}\times2\times(8^2-0)$

$K.E=64\ J$

Hence, The increase in it's kinetic energy one second after the start of motion is 64 J.

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