Question:medium

A body of mass \(2\) kg changes its velocity from \((3\hat i-4\hat j)\,m/s\) to \((6\hat j+2\hat k)\,m/s\). What is the change in kinetic energy of the body?

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For vector velocity, first calculate \(v^2\) by squaring and adding components.
  • \(15\,J\)
  • \(12\,J\)
  • \(18\,J\)
  • \(20\,J\)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
The change in kinetic energy is given by the difference between the final kinetic energy and the initial kinetic energy. The kinetic energy of a body is determined by its mass and the square of its speed (magnitude of velocity).
Step 2: Key Formula or Approach:
1. Kinetic Energy (KE) is given by $KE = \frac{1}{2}mv^2$, where $v$ is the speed. 2. The change in kinetic energy is $\Delta KE = KE_{final} - KE_{initial} = \frac{1}{2}mv_{final}^2 - \frac{1}{2}mv_{initial}^2$. 3. For a velocity vector $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$, the square of the speed is $v^2 = |\vec{v}|^2 = v_x^2 + v_y^2 + v_z^2$.
Step 3: Detailed Explanation:
Given: - Mass, $m = 2$ Kg. - Initial velocity, $\vec{v}_{initial} = 3\hat{i} - 4\hat{j}$ m/s. - Final velocity, $\vec{v}_{final} = 6\hat{j} + 2\hat{k}$ m/s. First, calculate the square of the initial speed: \[ v_{initial}^2 = |\vec{v}_{initial}|^2 = (3)^2 + (-4)^2 + (0)^2 = 9 + 16 = 25 \, (\text{m/s})^2 \] Calculate the initial kinetic energy: \[ KE_{initial} = \frac{1}{2}mv_{initial}^2 = \frac{1}{2}(2)(25) = 25 \text{ J} \] Next, calculate the square of the final speed: \[ v_{final}^2 = |\vec{v}_{final}|^2 = (0)^2 + (6)^2 + (2)^2 = 36 + 4 = 40 \, (\text{m/s})^2 \] Calculate the final kinetic energy: \[ KE_{final} = \frac{1}{2}mv_{final}^2 = \frac{1}{2}(2)(40) = 40 \text{ J} \] Finally, calculate the change in kinetic energy: \[ \Delta KE = KE_{final} - KE_{initial} = 40 \text{ J} - 25 \text{ J} = 15 \text{ J} \] Step 4: Final Answer:
The change in kinetic energy of the body is 15 J. Therefore, option (A) is correct.
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