Question

A force acts on a 2 kg object so that its position is given as a function of time as x = 3t² + 5. What is the work done by this force in first 5 seconds?
(A) 850 J (B) 950 J
(C) 875 J (D) 900 J

Answer 1

Answer:

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Explanation:

Answer 2

The work done by force in first 5 seconds is 900 J.

Option (D) is correct

Explanation:

Given :

The mass of object $m =$ 2 kg.

The position as a function of time  $x(t) = 3t^{2} +5$.

The work done by the force is equal to the change in kinetic energy.

⇒  $W = KE_{f} - KE_{i}$

And $KE = \frac{1}{2} mv^{2}$.

Now we have to find velocity from the position equation.

 $v = \frac{dx}{dt}$

 $v = 6t$

Given in question work done in first 5 sec. so put $t = 0$ and $t = 5$ in velocity equation.

$v_{i} = 0$ and $v_{f} = 30$

Put the value of velocity in work equation.

⇒ $W = \frac{1}{2} \times 2 \times (30)^{2}$

   $W = 900$ J

Thus, the work done by force in first 5 seconds is 900 J.