Home
Class 12
PHYSICS
If a constant force acts on a particle, ...

If a constant force acts on a particle, its acceleration will

A

remain constant

B

gradually decrease

C

gradully increase

D

be undefined.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, "If a constant force acts on a particle, its acceleration will," we can follow these steps: ### Step 1: Understand the relationship between force, mass, and acceleration According to Newton's second law of motion, the relationship between force (F), mass (m), and acceleration (a) is given by the equation: \[ F = m \cdot a \] This means that the force acting on an object is equal to the mass of the object multiplied by its acceleration. ### Step 2: Analyze the given conditions The problem states that a constant force is acting on the particle. This implies that the value of F is constant. Additionally, we need to consider the mass of the particle. If the velocity of the particle is much less than the speed of light (non-relativistic speeds), the mass of the particle can be considered constant. ### Step 3: Determine the effect of constant force on acceleration Since both the force (F) and mass (m) are constant (under non-relativistic conditions), we can rearrange the equation to find acceleration: \[ a = \frac{F}{m} \] Since F is constant and m is constant, it follows that the acceleration (a) will also remain constant. ### Step 4: Consider relativistic effects However, if the velocity of the particle increases and approaches a significant fraction of the speed of light, relativistic effects must be considered. In relativity, the mass of the particle increases with velocity according to the equation: \[ m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \] where \( m_0 \) is the rest mass, \( v \) is the velocity of the particle, and \( c \) is the speed of light. ### Step 5: Analyze the implications of increasing velocity As the velocity \( v \) approaches the speed of light \( c \), the denominator in the mass equation decreases, causing the mass \( m \) to increase. Since the force \( F \) remains constant, but the mass \( m \) is increasing, the acceleration \( a \) must decrease: \[ a = \frac{F}{m} \] As \( m \) increases, \( a \) decreases. ### Conclusion Initially, if the particle's velocity is much less than the speed of light, the acceleration remains constant. However, as the velocity increases and becomes comparable to the speed of light, the acceleration decreases. Therefore, the correct answer to the question is that the acceleration will gradually decrease. ### Final Answer The acceleration will gradually decrease.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • THE SPECIAL THEORY OF RELATIVITY

    HC VERMA ENGLISH|Exercise Objective 2|1 Videos
  • THE SPECIAL THEORY OF RELATIVITY

    HC VERMA ENGLISH|Exercise Exercise|28 Videos
  • THE SPECIAL THEORY OF RELATIVITY

    HC VERMA ENGLISH|Exercise worked out example|8 Videos
  • THE NUCLEOUS

    HC VERMA ENGLISH|Exercise Short answer|12 Videos
  • THERMAL AND CHEMICAL EFFECT OF ELECTRIC CURRENT

    HC VERMA ENGLISH|Exercise Short Question|9 Videos

Similar Questions

Explore conceptually related problems

A constant force acts on a particle and its displacement x (in cm) is related to the time t (in s) by the equation t=sqrtx +3 , when the velocity of the particle is zero, its displacement ( in cm) is

The momentum of a particle is P = A + Bt^(2) . Where , A and B are constant perpendicular vectors . The force acting on the particle when its acceleration is at 45^(@) with its velocity is (a) 2 sqrt((A)/(B)) B (b)2B (c)zero (d)2A

When a force F acts on a particle of mass m, the acceleration of particle becomes a. now if two forces of magnitude 3F and 4F acts on the particle simultaneously as shown in figure, then the acceleration of the particle is

A force P acts on a particle so as to accelerate it from rest to certain velocity. If P is replaced then by Q which decelerates it to rest.

A particle at rest on a frictionless table is acted upon by a horizontal force which is constant in magnitude and direction. A graph is plotted of the work done on the particle W, against the speed of the particle v. If there are no frictional forces acting on the particle, the graph will look like

According to Newton.s second law of motion, resultant force on a particle is in the direction of acceleration of the particle. Two forcce F_(1) and F_(2) are acting on a particle as shown in the figure. The acceleration of the particle is along X-axis. FInd the value of F_(1) (in newton)

A particle of mass 70 g, moving at 50 cm/s , is acted upon by a variable force opposite to its direction of motion. The force F is shown as a function of time t (a) its speed will be 50 cm/s after the force stops acting (b) its direction of motion will reverse (c) its average acceleration will be 1 m//s^2 during the interval in which the force acts its average acceleration will be 10 m//s^2 during the interval in which the force acts

A force F_1 acts on a particle so as to accelerate it from rest to a velocity v. The force F_1 is then replaced by F_2 which decelerates it to rest.

A force F_1 acts on a particle so as to accelerate it from rest to a velocity v. The force F_1 is then replaced by F_2 which decelerates it to rest

Under the action of a constant force, A particle is experiencing a constant acceleration. The power is