Question

A $3 kg$ object has initial velocity $\left( {6\hat i - 2\hat j} \right)\,{\text{m/s}}$. The total work done on the object if its velocity changes to $\left( {8\hat i + 4\hat j} \right)\,{\text{m/s}}$ is:
A. $216 J$
B. $44 J$
C. $60 J$
D. $120 J$

Answer 1

Determine the magnitude of the initial and final velocity. Use the Work-energy theorem to relate the change in kinetic energy with the work done.

Formula used:
The kinetic energy of the body of mass m moving with velocity v is,
$K.E = \dfrac{1}{2}m{v^2}$
Work-energy theorem,
$W = \Delta K.E$
Here, $\Delta K.E$ is the change in kinetic energy.

Complete step by step answer:
We have given the initial velocity of the object ${v_i} = \left( {6\hat i - 2\hat j} \right)\,{\text{m/s}}$ and final velocity ${v_f} = \left( {8\hat i + 4\hat j} \right)\,{\text{m/s}}$. The mass of the object is $m = 3\,{\text{kg}}$.
We know from the work-energy theorem, the work done is equal to the change in kinetic energy of the body. Therefore, we can write,
$W = {K_f} - {K_i}$
$\Rightarrow W = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2$ …… (1)
Let’s determine the magnitude of the final velocity as follows,
$v_f^2 = v_{xf}^2 + v_{yf}^2$
Here, ${v_{xf}}$ is the x-component of the final velocity and ${v_{yf}}$ is the y-component of final velocity.
Substituting 8 m/s for ${v_{xf}}$ and 4 m/s for ${v_{yf}}$ in the above equation, we get,
$v_f^2 = {\left( 8 \right)^2} + {\left( 4 \right)^2}$
$\Rightarrow {{v_f}^2} = 80$ ……. (2)
Let’s determine the magnitude of the initial velocity as follows,
${{v_i}^2} = {{v_{xi}}^2} + {{v_{yi}}^2}$
Here, ${v_{xi}}$ is the x-component of the initial velocity and ${v_{yi}}$ is the y-component of initial velocity.
Substituting 6 m/s for ${v_{xi}}$ and $- 2$ m/s for ${v_{yf}}$ in the above equation, we get,
$v_i^2 = {\left( 6 \right)^2} + {\left( { - 2} \right)^2}$
$\Rightarrow v_i^2 = 40$ ……. (3)

Using equation (2) and (3) in equation (1) and substituting the value of mass in the same to calculate the work done as follows,
$W = \dfrac{1}{2}\left( 3 \right)\left( {80} \right) - \dfrac{1}{2}\left( 3 \right)\left( {40} \right)$
$\therefore W = 60\,{\text{J}}$

Therefore, the work done is 60 J.So, the correct answer is option C.

Note: To calculate the kinetic energy, students often directly take the square of velocities ${v_i} = \left( {6\hat i - 2\hat j} \right)\,{\text{m/s}}$and ${v_f} = \left( {8\hat i + 4\hat j} \right)\,{\text{m/s}}$. This is an incorrect way to solve these types of questions. Always find the magnitude of the velocity using the expression, ${v^2} = v_x^2 + v_y^2$.