Question

Force acting on a particle moving with a constant speed on a straight-line path is
A. Always zero
B. Need not be zero
C. Always non zero
D. Cannot be concluded

Answer 1

Hint: In this question, two key things are given. First, the particle moves with a constant speed and second that the particle moves on a straight path. Since the motion is a straight line and the direction of motion is not changing, velocity is equal to the speed. Acceleration is defined as the change in velocity. We will calculate the acceleration and substitute in the force equation $F = ma$ to get the answer.

Complete answer:
Force acting on a particle is given by $F = ma$ where F is the force acting on the particle, m is the mass of the particle and a is the acceleration of the particle.
The mass of the particle would remain constant. So, the force acting on the particle would be determined by its acceleration.
Acceleration is the change in the velocity of the particle in a given time interval. Velocity in magnitude is equal to the speed of the particle. The only difference is that a direction is specified along with magnitude for the velocity.
In this question, since the particle moves on a straight-line path, there is no change in the direction of motion. Hence, the velocity is the same as the speed of the particle.
Also, it is given that the particle is moving on a straight-line path with a constant speed. This means that the speed of the particle is not changing which in turn means that the velocity is also not changing.
Thus, we can say that the change in the velocity is zero. Therefore, the acceleration is also zero.
Substituting in the force equation, we get
$F = m \times 0$
$ \Rightarrow F = 0$
Hence, option (A) is the correct answer.

Note:
 It is not always that the motion with a constant speed has no change of direction associated with it. Consider a stone tied to a string being rotated in air. Here the speed would be constant. But since the stone moves in a circular path, there will be a direction change at every point. And so, some acceleration and hence some force must be acting in such a case.