A particle is moving in a circular path of radius r under the action of a force F . If at an instant velocity of particle is v, and speed of particle is increasing, then

Question

A particle is moving in a circular path of radius r under the action of a force F . If at an instant velocity of particle is v, and speed of particle is increasing, then (A)  vecF ⋅ vecv=0 (B)  vecF ⋅ vecv > 0 (C)  vecF ⋅ vecv < 0 (D)  vecF ⋅ vecv ≥ 0

Tardigrade Admin · Accepted Answer

Net force will be in the direction of net acceleration. Here accelerations are of two types (i) Centripetal (ii) Tangential θ<90° always ⇒ vecF ⋅ vecv>0

Q. A particle is moving in a circular path of radius $r$ under the action of a force $F .$ If at an instant velocity of particle is $v$ , and speed of particle is increasing, then

$F \cdot v = 0$

$F \cdot v > 0$

$F \cdot v < 0$

$F \cdot v \geq 0$

Net force will be in the direction of net acceleration.
Here accelerations are of two types
(i) Centripetal
(ii) Tangential
$θ < 9 0^{\circ}$ always
$\Rightarrow F \cdot v > 0$

Q. A particle is moving in a circular path of radius r under the action of a force F. If at an instant velocity of particle is v, and speed of particle is increasing, then

F⋅v=0

F⋅v>0

F⋅v<0

F⋅v≥0