The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
1 answer:
Answer:
The y-component of the electric force on this charge is
Explanation:
<u>Given:</u>
- Electric field in the region,
- Charge placed into the region,
where, are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,
Thus, the y-component of the electric force on this charge is
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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: