Question

Under the action of a force, a 1 kg body moves, such that its position x as a function of time t is given by 3 , 2 t x  where x is in meter and t is in second. The work done by the force in first 3 second is

Answer 1

$\huge{\underline{\underline{\mathfrak{Correct\: Question:-}}}}$

Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x= t ⅔ , where x is in metre and t in second. The work done by the force in first two seconds is ?

$\huge{\underline{\underline{\mathfrak{Answer:-}}}}$

Work done by force is $\frac{16}{9}$ joules.

$\huge{\underline{\underline{\mathfrak{Explainationtion:-}}}}$

$\large\bigstar{\underline{\mathfrak{Given:-}}}$

• Mass of the object, m = 2 kg

Its position as a function of time t is given by:

t $= \frac{ {t}^{2} }{3}$

Velocity, v = $\frac{dx}{dt}$ = $\dfrac{2t}{3}$

Where:

• x is in meters
• t is in seconds

We need to find the work done by the force in first two seconds. The work done by an object is given by:

$W = \frac{1}{2}m( {v}^{2} - {u}^{2}) \implies \: eq \: (1)$

Where:

• v is the final velocity.
• u is the initial velocity.

At, t = 0, u = 0

At, t = 2 s, v = $\dfrac{4}{3}$

Put the values of u and v in equation (1), we get:

$W = \frac{1}{2} \times 2[( {\frac{4}{3} })^{2} - 0)]$

$W = \frac{16}{9}J$

So, the work done by the force in first two seconds is $\dfrac{16}{9}$ Joules.

Hence, this is the required solution.

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Answer 2

correct Question :

Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by $x = \frac{t {}^{2} }{3}$ where x is in metre and t in second. The work done by the force in first two

seconds is

(1) 1600

(2) 160

(3) 160

(4) 16/9

Theory :

work energy theorem:

${\purple{\boxed{\large{\bold{Work \: Done = Change \: Kinetic \: energy }}}}}$

Solution :

given: mass of the body = 2 kg

position as a function of time is given by : $x = \frac{t {}^{2} }{3}$

differnatiate with respect to x

⇒$velocity , \frac{dx}{dt} = \frac{2t}{3}$

when t = 0 sec

$v _{1} = 0 \: ms {}^{ - 1}$

when t = 2 seconds

$v_{2} = \frac{2 \times 2}{3} = \frac{4}{3} m s {}^{ - 1}$

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Now apply work energy theorem:

$w = \delta \: kitenic \: energy \:$

$w = k _{2} - k _{1}$

$w = \frac{1}{2} mv {}^{2} _{2} - \frac{1}{2} mv {}^{2}_{1}$

$w = \frac{1}{2} m(v {}^{2} _{2} - v {}^{2} _{1})$

Now put the values of v1 ,V2 and m

$\implies \: w = \frac{1}{2} \times 2(( \frac{4}{3}) {}^{2} - 0)$

$\implies \: w = \frac{16}{9} J$

${\purple{\boxed{\large{\bold{Work \: Done =\frac{16}{9 } Jolues}}}}}$

So, the work done by the force in first two seconds 16/9 joules .