Answer:
Explanation:
By using the Newton second law and the position equation for a simple harmonic motion we have
where a is the acceleration, w is the angular frecuency and \phi is the phase constant. We can calculate w from the equation for the maximum acceleration
(a).
(b).
(c).
(d). The mecanical energy is the kinetic energy when the velocity is a maximum
Answer:
u= 20.09 m/s
Explanation:
Given that
m = 0.02 kg
M= 2 kg
h= 0.2 m
Lets take initial speed of bullet = u m/s
The final speed of the system will be zero.
From energy conservation
1/2 m u²+ 0 = 0+ (m+M) g h
m u²=2 (m+M) g h
By putting the values
0.02 x u² = 2 (0.02+2) x 10 x 0.2 ( take g=10 m/s²)
u= 20.09 m/s
Work done is given by product of force and displacement due to that force
So here we will have
here we know that
Now work done is given as
so it will do 16 J work to move the box
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