Question

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?

• A. 15 J
• B. 12 J
• C. 18 J
• D. 20 J

Hint:

According to the Work-Energy Theorem, the net work done on an object equals its change in kinetic energy. This calculation gives the net work done on the body to change its velocity. Remember to find the magnitude (speed) from the velocity vector before calculating KE.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the mass and the initial and final velocity vectors of a body. We need to calculate the change in its kinetic energy.

Step 2: Key Formula or Approach:
The change in kinetic energy (\(\Delta KE\)) is the final kinetic energy minus the initial kinetic energy.
\[ \Delta KE = KE_{final} - KE_{initial} \]
The kinetic energy is given by \(KE = \frac{1}{2}mv^2\), where \(v\) is the speed (magnitude of the velocity vector).

Step 3: Detailed Explanation:
The mass of the body is \(m = 2\) Kg.
Initial velocity, \(\vec{v}_i = 3\hat{i} - 4\hat{j}\).
Final velocity, \(\vec{v}_f = 6\hat{j} + 2\hat{k}\).
First, calculate the initial speed squared (\(v_i^2\)):
\[ v_i^2 = |\vec{v}_i|^2 = (3)^2 + (-4)^2 = 9 + 16 = 25 \, (\text{m/s})^2 \]
Now, calculate the initial kinetic energy (\(KE_i\)):
\[ KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(2)(25) = 25 \text{ J} \]
Next, calculate the final speed squared (\(v_f^2\)):
\[ v_f^2 = |\vec{v}_f|^2 = (6)^2 + (2)^2 = 36 + 4 = 40 \, (\text{m/s})^2 \]
Now, calculate the final kinetic energy (\(KE_f\)):
\[ KE_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(2)(40) = 40 \text{ J} \]
Finally, calculate the change in kinetic energy:
\[ \Delta KE = KE_f - KE_i = 40 \text{ J} - 25 \text{ J} = 15 \text{ J} \]

Step 4: Final Answer:
The change in kinetic energy of the body is 15 J.