First slide

Kinetic energy

Question

A particle moves in the x-y plane under the action of a force $\vec{F}$ such that the value of its linear momentum $\vec{P} at any time t is P_{x} = 2 \cos t,$
$P_{y} = 2 \sin t$ . The angle $θ$ between $\vec{F}$ and $\vec{P}$ at a given time t will be

Moderate

A

$90^{0}$
B

$0^{0}$
C

$180^{0}$
D

$30^{0}$

Solution

$P = \sqrt{{P_{x}}^{2} + {P_{y}}^{2}} = \sqrt{{(2 \cos t)}^{2} + {(2 \sin t)}^{2}} = 2$
If m be the mass of the body, then
Kinetic energy = $\frac{P^{2}}{2 m} = \frac{{(2)}^{2}}{2 m} = \frac{2}{m}$
Since kinetic energy does not change with time, both work done and power are zero.
Now Power = Fv $\cos θ = 0$
As $F \neq 0, v \neq 0 ∴ \cos θ or θ = 90^{0}$
As direction of $\vec{P} is same that \vec{v} (∴ \vec{P} = m \vec{v}), hence angle between \vec{F} and \vec{P} is equal to 90^{0}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App