Question

A 3kg 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) $216{\text{ J}}$
B) $44{\text{ J}}$
C) ${\text{60 J}}$
D) ${\text{120 J}}$

Answer 1

In the given problem, the velocity of the object is changing, and we have to find total work done on the object, to solve this problem we will use the WORK-ENERGY Theorem. According to the WORK-ENERGY Theorem “total work done of the object equals the change in kinetic energy of the object.

Complete step by step answer:
Since the velocity of the object is given in terms of the vector. So, we have to calculate the magnitude of the velocity at the initial and final positions. Let’s assume the magnitude of the initial velocity is ${V_1}$ and the magnitude of the final velocity is ${V_2}$.
The given initial velocity is ${\vec V_1} = \left( {6\hat i - 2\hat j} \right){\text{ m/s}}$.
Here, the velocity along the horizontal direction is ${v_x} = 6 m/s$ and
The velocity along the vertical direction is ${v_y} = -2m/sec$
Now, calculate the magnitude of the initial velocity as follows:
${V_1} = \sqrt {v_x^2 + v_y^2}$
$\Rightarrow {V_1} = \sqrt {{6^2} + {{\left( { - 2} \right)}^2}}$
$\Rightarrow {V_1} = \sqrt {36 + 4}$
$\Rightarrow {V_1} = 6.32{\text{ m/s}}$
The given initial velocity is ${\vec V_2} = \left( {8\hat i + 4\hat j} \right){\text{ m/s}}$.
Here, the velocity along the horizontal direction is $v_x = 8m/s$
and the velocity along the vertical direction is $v_y =4 m/s$
Similarly, calculate the magnitude of the final velocity as follows
$\Rightarrow {V_2} = \sqrt {v_x^2 + v_y^2}$
$\Rightarrow {V_2} = \sqrt {{8^2} + {4^2}}$
$\Rightarrow {V_2} = \sqrt {64 + 16}$
$\Rightarrow {V_2} = 8.94{\text{ m/s}}$
Let’s assume the initial kinetic energy is ${K_1}$ and the final kinetic energy is ${K_2}$.
Calculate the initial kinetic energy of the object as follows:
$\Rightarrow {K_1} = \dfrac{1}{2}mV_1^2$
$\Rightarrow {K_1} = \dfrac{1}{2} \times 3 \times {6.32^2}$
$\Rightarrow {K_1} = 59.91{\text{ J}}$
Similarly, calculate the final kinetic energy of the object as follows:
$\Rightarrow {K_2} = \dfrac{1}{2}mV_2^2$
$\Rightarrow {K_2} = \dfrac{1}{2} \times 3 \times {8.94^2}$
$\Rightarrow {K_2} = 119.88{\text{ J}}$
Now, use the WORK ENERGY Theorem, according to the WORK-ENERGY Theorem “total work done of the object equals the change in kinetic energy of the object”.
${\text{total work done}} = {\text{change in kinetic energy}}$
$\Rightarrow {W_{total}} = {K_2} - {K_1}$
$\Rightarrow {W_{total}} = {\text{119}}{\text{.88 J}} - 59.91{\text{ J}}$
$\Rightarrow {W_{total}} = 59.97{\text{ J}}$
$\Rightarrow {W_{total}} \simeq 60{\text{ J}}$

$\therefore$ The total work done on the object is $60{\text{ J}}$. Therefore, the correct option is (C).

Note:
Please take care while calculating the kinetic energy of the object. First, calculate the magnitude of the velocity at the initial and final position and then apply the kinetic energy equation to get kinetic energy. While applying the WORK ENERGY theorem, the initial kinetic energy of the object should be subtracted from the final kinetic energy.