Answer:
F3 is the equilibrant force equal in magnitude to the resultant between F1 and F2 but opposite in direction to it.
Explanation:
Given the diagram, the force F3 to make the body remain at rest or in equilibrium is the equilibrant force, this force is equal in magnitude to the resultant Fr between F1 and F2 but opposite in direction to it.
See attachment for diagram of forces.
The resultant force;
Fr =√ (F1)² + (F2)²...(1) [diagram a]
Therefore the length of F3 is Fr
F3 = -Fr
The direction (diagram b) of the resultant force Fr is given by
∆ = acrtan[(F1/F2)]
The direction of F3 is [90 + arctan(F1/F2)]
As seen in the diagram (d), the location of the force is in the fourth quadrant.
Answer:
W = 2.74 J
Explanation:
The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.
This is the electrostatic equivalent of the work-energy theorem.
where the potential energy is defined as follows
Let's first calculate the distance 'r' for both positions.
Now, we can calculate the potential energies for both positions.
Finally, the total work done on the moving particle can be calculated.
Answer:
The kinetic energy of the system after the collision is 9 J.
Explanation:
It is given that,
Mass of object 1, m₁ = 3 kg
Speed of object 1, v₁ = 2 m/s
Mass of object 2, m₂ = 6 kg
Speed of object 2, v₂ = -1 m/s (it is moving in left)
Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,
E = 9 J
So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.