Kinetic Energy of Block Subjected to Time Dependent Force

To find the kinetic energy of the block after $t = 4.5\text{ s}$, we can use the Impulse-Momentum Theorem, which states that the total impulse given to an object is equal to its change in momentum:

$\Delta p = J = \int_{0}^{t} F(t) \, dt$

Step 1: Determine the equation for $F(t)$

From the given $F-t$ graph, the force decreases linearly with time.

• At $t = 0\text{ s}$, $F(0) = 4\text{ N}$
• At $t = 3\text{ s}$, $F(3) = 0\text{ N}$

The slope of the line is:
$m = \frac{0 - 4}{3 - 0} = -\frac{4}{3}\text{ N/s}$

Thus, the force as a function of time $t$ is:
$F(t) = 4 - \frac{4}{3}t$

At $t = 4.5\text{ s}$, the force becomes:
$F(4.5) = 4 - \frac{4}{3}(4.5) = 4 - 6 = -2\text{ N}$

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Step 2: Calculate the net impulse up to $t = 4.5\text{ s}$

The net impulse $J$ is given by the area under the $F-t$ curve from $t = 0$ to $t = 4.5\text{ s}$:

$J = \int_{0}^{4.5} \left(4 - \frac{4}{3}t\right) dt$

Evaluating the integral:
$J = \left[ 4t - \frac{2}{3}t^2 \right]_{0}^{4.5}$
$J = 4(4.5) - \frac{2}{3}(4.5)^2$
$J = 18 - \frac{2}{3}(20.25) = 18 - 13.5 = 4.5\text{ N}\cdot\text{s}$

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Step 3: Determine final momentum and kinetic energy

Since the block is initially at rest ($p_i = 0$), the final momentum at $t = 4.5\text{ s}$ is:
$p_f = J = 4.5\text{ kg}\cdot\text{m/s}$

Given the mass of the block $m = 2\text{ kg}$, the kinetic energy $K$ is:
$K = \frac{p_f^2}{2m}$
$K = \frac{(4.5)^2}{2 \times 2} = \frac{20.25}{4} = 5.0625\text{ J} \approx 5.06\text{ J}$

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Conclusion

The kinetic energy of the block after $4.5\text{ seconds}$ is $5.06\text{ J}$, which corresponds to Option C.