Question

A 12.0N force with a fixed orientation does work on a

Answer 1

Answer:

(a) $\theta=62.31^{\circ}$

(b) $\theta=117.68^{\circ}$

Explanation:

It is given that,

Force acting on the particle, F = 12 N

Displacement of the particle, $d=(2.00i -4.00j+3.00k)\ m$

Magnitude of displacement, $d=\sqrt{2^2+4^2+3^2}= 5.38\ m$

(a) If the change in the kinetic energy of the particle is +30 J. The work done by the particle is given by :

$W=Fd\ cos\theta$

$\theta$ is the angle between force and the displacement

According to work energy theorem, the charge in kinetic energy of the particle is equal to the work done.

So,

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{+30\ J}{12\times 5.38}$

$\theta=62.31^{\circ}$

(b) If the change in the kinetic energy of the particle is (-30) J. The work done by the particle is given by :

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{-30\ J}{12\times 5.38}$

$\theta=117.68^{\circ}$

Hence, this is the required solution.