The position (x) of a particle of mass 2 kg moving along x-axis at time t is given by `x=(2t^(3))` metre. Find the work done by force acting on it in time interval t=0 to t=2 is :-

Options

A. 576 J
B. 584 J
C. 623 J
D. None

Answer

To solve the problem, we need to find the work done by the force acting on a particle of mass 2 kg, given its position as a function of time. The position is given by:

\[ x(t) = 2t^3 \text{ meters} \]

**Step 1: Find the velocity of the particle.**

The velocity \( v(t) \) is the derivative of the position \( x(t) \) with respect to time \( t \):

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2t^3) = 6t^2 \text{ m/s} \]

**Step 2: Find the acceleration of the particle.**

The acceleration \( a(t) \) is the derivative of the velocity \( v(t) \):

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2) = 12t \text{ m/s}^2 \]

**Step 3: Calculate the kinetic energy at \( t = 0 \) and \( t = 2 \).**

The kinetic energy \( KE \) is given by the formula:

\[ KE = \frac{1}{2} mv^2 \]

- At \( t = 0 \):

\[ v(0) = 6(0)^2 = 0 \text{ m/s} \]
\[ KE(0) = \frac{1}{2} \times 2 \times (0)^2 = 0 \text{ J} \]

- At \( t = 2 \):

\[ v(2) = 6(2)^2 = 6 \times 4 = 24 \text{ m/s} \]
\[ KE(2) = \frac{1}{2} \times 2 \times (24)^2 = 1 \times 576 = 576 \text{ J} \]

**Step 4: Calculate the work done by the force.**

According to the work-energy theorem, the work done \( W \) is equal to the change in kinetic energy:

\[ W = KE(2) - KE(0) = 576 \text{ J} - 0 \text{ J} = 576 \text{ J} \]

Thus, the work done by the force acting on the particle in the time interval from \( t = 0 \) to \( t = 2 \) seconds is:

\[ \boxed{576 \text{ J}} \]

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