Question

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

• A. \(15\text{ J}\)
• B. \(12\text{ J}\)
• C. \(18\text{ J}\)
• D. \(20\text{ J}\)

Hint:

For vector velocity, first calculate the square of speed using \(v^2=v_x^2+v_y^2+v_z^2\), then use \(K=\frac{1}{2}mv^2\).

Answer 1

The Correct Option is A

We are given the mass of the body: \[ m=2\text{ kg}. \] Initial velocity is: \[ \vec{u}=3\hat{i}-4\hat{j}. \] Final velocity is: \[ \vec{v}=6\hat{j}+2\hat{k}. \] The kinetic energy of a body is: \[ K=\frac{1}{2}mv^2. \] Here \(v^2\) means the square of the magnitude of velocity vector. First, find the square of initial speed: \[ u^2=(3)^2+(-4)^2. \] \[ u^2=9+16. \] \[ u^2=25. \] Now find the square of final speed: \[ v^2=(6)^2+(2)^2. \] \[ v^2=36+4. \] \[ v^2=40. \] Change in kinetic energy is: \[ \Delta K=\frac{1}{2}m(v^2-u^2). \] Substitute the values: \[ \Delta K=\frac{1}{2}(2)(40-25). \] \[ \Delta K=1\times 15. \] \[ \Delta K=15\text{ J}. \] Therefore, the change in kinetic energy is: \[ 15\text{ J}. \]