Particles q1 = +8.0 UC, 92 = +3.5 uc, and
2 answers:
Answer:
F_total = 29.4 N, directed to the right of particle 2
Explanation:
We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.
Let's use Coulomb's law to calculate each force
F =
particles 1 and 2
q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m
F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²
F₁₂ = 2.59 10¹ N
Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.
particles 2 and 3
q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m
we calculate
F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²
F₂₃ = 3.5 N
as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2
Now we add the forces as vectors
F_total = ∑ F = F₁₂ + F₂₃
F_total = 25.2 +3.5
F_total = 29.4 N
directed to the right of particle 2
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Answer:
v = 0.41 m/s
Explanation:
- In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
- At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
- So, we can write the following general equation, taking the initial and final values of the energies:
- Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
- ⇒ Kf = 1/2*m*vf² (2)
- The change in the potential energy, can be written as follows:
where k = force constant = 815 N/m
xf = final displacement of the block = 0.01 m (taking as x=0 the position
for the spring at equilibrium)
x₀ = initial displacement of the block = 0.03 m
- Regarding the work done by the force of friction, it can be written as follows:
where μk = coefficient of kinettic friction, Fn = normal force, and Δx =
horizontal displacement.
- Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
- Fn = Fg= m*g (5)
- Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:
- Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get: