Question

A $3kg$ object has initial velocity $(6\hat{i}-2\hat{j})m/s$. The total work done on the object if its velocity changes to $(8\hat{i}+4\hat{j})m/s$ is :

• A. $216J$
• B. $44J$
• C. $60J$
• D. $120J$

Answer 1

Answer: (C)

$60J$

The net workdone on the object is equal to the change in the kinetic energy of the object
$W_{net}=K_f-K_i=△K$
Kinetic energy $K=\frac{1}{2}m{v}^2$
Velocity $v^2=v_x^2+v_y^2$
$=(36+4)m/s^2$
$v^2=40m/s^2$
$K_i=\frac{1}{2}\times 3\times 40=60J$
Again velocity $v=\sqrt{8^2+4^2}$
$v=\sqrt{64+16}=\sqrt{80}$
Kinetic energy $K_f=\left(\frac{1}{2}\right)\left(3\right)\times 80m/s^2=120J$
From work energy theorem
$W_{net}=K_f-K_i$
$=120J-60J=60J$