Question

A mysterious force acts on all particles along a particular line and always points towards a particular point P on the line. The magnitude of the force on a particle increases as the cube of the distance from that point, that is, F∝ r3, if the distance from the P to the position of the particle is r. It has been determined that the constant of proportionality is 0.45 N/m3, i.e. the magnitude of the force on a particle can be written as 0.45r3, when the particle is at a distance r from the force center.

Answer 1

The magnitude of the force on a particle increases as the cube of the distance from a specific point, and can be represented by the equation F = br³. To find the potential energy of a particle at a certain distance from the force center, we can integrate the force equation over the distance and calculate the negative work done by the force.

Step-by-step explanation:

In this scenario, we are dealing with a force that acts on particles along a particular line and always points towards a specific point P on the line. The magnitude of the force on a particle increases as the cube of the distance from that point. The force can be represented by the equation F = br³, where F is the magnitude of the force, b is the proportionality constant, and r is the distance from point P to the position of the particle.

To find the potential energy of a particle subjected to this force when the particle is at a distance D from P, we can integrate the force over the distance from P to D. The potential energy can be calculated as the negative of the work done by the force.

Using the equation F = br³, we can integrate this equation over the distance from P to D. The work done by the force is given by the equation W = -∫F dr, where r is the distance variable. By substituting F = br³ into the equation, we can solve the integral to find the potential energy.

Answer 2

`PE = (0.45 r^4)/(4)`

Step-by-step explanation:

given,

proportionality constant = 0.45

F∝ r³

F = k r³

using conservation of energy change in potential energy is equal to work done.

d (PE) = F. dr

d (PE) = k r³. dr

integrating both side

`\int {d (PE)} = k \int {r^3} dr`

`PE = (k r^4)/(4) + C`

given that at r = 0; PE = 0

`0 = (k 0^4)/(4) + C`

C = 0

`PE = (k r^4)/(4)`

`PE = (0.45 r^4)/(4)`