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

Hint: 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, \[{v^2} = v_x^2 + v_y^2\].