A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
1 answer:
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:
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Answer:
There is a loss of fluid in the container of 0.475L
Explanation:
To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.
The formula that describes this thermal expansion process is given by:
Where,
Change in volume
Initial Volume
Change in temperature
coefficient of volume expansion (Coefficient of copper and of the liquid for this case)
There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,
Where,
= Change in the volume of liquid
= Change in the volume of copper
Then replacing with the previous equation we have:
Our values are given as,
Thermal expansion coefficient for copper and the liquid to 20°C is
Replacing we have that,
Therefore there is a loss of fluid in the container of 0.475L
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]