Question

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

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

Answer 1

Answer: (A)

The net workdone on the object is equal to the change in the kinetic energy of the object
$W_{net}=K_f-K_i=\Delta K$
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 $K_f=\left(\frac{1}{2}\right)(3)\times 80m/s^2=120J$
From work energy theorem
$W_{\text{net }}=K_f-K_i$
$=120J-60J=60J$