A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
1 answer:
Answer:
a) k = 200 N/m
b) E = 4 J
c) Δx = 6.3 cm
Explanation:
a)
- In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:
- where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
- Solving for k:
b)
- Assuming no friction present, total mechanical energy mus keep constant.
- When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:
- Replacing k and Δx by their values, we get:
c)
- When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
- We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
- So, we can write the following expression:
- Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:
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Answer:
(b) EAST
Explanation:
you can assume that the magnetic field points rightward, that is, in the positive x direction (NORTH). Furthermore, you can assume that the direction of the motion of the electron is in the positive y direction. Hence, you have:
You use the Lorentz formula to known which is the direction of the magnetic force over the electron:
which implies the cross product between the unitary vecors j and i, that is
(WEST)
However, the minus sign of the charge of the electron changes the direction 180°. Hence, the direction is k. That is, to the EAST
Answer:
a) E = -4 10² N / C , b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰
Explanation:
For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball
X axis
-
= m a
Axis y
- W = 0
Initially the system is in equilibrium, so zero acceleration
Fe =
T_{y} = W
Let us search with trigonometry the components of the tendency
cos θ = T_{y} / T
sin θ = / T
T_{y} = cos θ
= T sin θ
We replace
q E = T sin θ
mg = T cosθ
a) the electric force is
= q E
E = / q
E = -0.032 / 80 10⁻⁶
E = -4 10² N / C
b) the distance to this point can be found by dividing the two equations
q E / mg = tan θ
θ = tan⁻¹ qE / mg
Let's calculate
θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)
θ = tan⁻¹ 0.3265
θ = 18 ⁰
sin 18 = x/0.30
x =0.30 sin 18
x = 0.093 m
c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations
X axis
= m aₓ
aₓ = q E / m
aₓ = 80 10⁻⁶ 4 10² / 0.01
aₓ = 3.2 m / s²
Axis y
W = m
a_{y} = g
a_{y} = 9.8 m/s²
The total acceleration is can be found using Pythagoras' theorem
a = √ aₓ² + a_{y}²
a = √ 3.2² + 9.8²
a = 10.31 m / s²
The Angle meet him with trigonometry
tan θ = a_{y} / aₓ
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-9.8) / 3.2
θ = -71.9⁰
Movement is two-dimensional type with acceleration in both axes