Newton's _____ law explains why my hands hurt when I clap loudly
2 answers:
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a) See free-body diagram in attachment
b) The book is stationary in the vertical direction
c) The net horizontal force is 35 N in the forward direction
d) The net force on the book is 35 N in the forward horizontal direction
e) The acceleration is in the forward direction
Explanation:
a)
The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.
From the diagram of this book, we see there are 4 forces acting on the book:
- The applied force, F = 50 N, pushing forward in the horizontal direction
- The frictional force, , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)
- The weight of the book, , where m is the mass of the book and
is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:
- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book
b)
The book is in equilibrium in the vertical direction, therefore there is no motion.
In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is
where a is the acceleration; however, since N = W, this becomes
And since the book is initially at rest on the desk, this means that there is no motion.
c)
We said there are two forces acting in the horizontal direction:
- The applied force, F = 50 N, forward
- The frictional force, , backward
Since they act along the same line, we can calculate their resultant as
and therefore the net force is 35 N in the forward direction.
d)
The net force is obtained as the resultant of the net forces in the horizontal and vertical direction. However, we have:
- The net force in the horizontal direction is 35 N
- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction
Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.
e)
The acceleration of the book can be calculated by using Newton's second law:
where
is the net force
m is the mass
a is the acceleration
Here we have:
(in the forward direction)
m = 4 kg
Therefore, the acceleration is
(forward)
Learn more about forces, weight and Newton's second law:
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Answer:
(a) the final velocity of the cart is 0.857 m/s
(b) the impulse experienced by the package is 21.43 kg.m/s
(c) the fraction of the initial energy lost is 0.71
Explanation:
Given;
mass of the package, m₁ = 10 kg
mass of the cart, m₂ = 25 kg
initial velocity of the package, u₁ = 3 m/s
initial velocity of the cart, u₂ = 0
let the final velocity of the cart = v
(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
10 x 3 + 25 x 0 = v(10 + 25)
30 = 35v
v = 30 / 35
v = 0.857 m/s
(b) the impulse experienced by the package;
The impulse = change in momentum of the package
J = ΔP = m₁v - m₁u₁
J = m₁(v - u₁)
J = 10(0.857 - 3)
J = -21.43 kg.m/s
the magnitude of the impulse experienced by the package = 21.43 kg.m/s
(c)
the initial kinetic energy of the package is calculated as;
the final kinetic energy of the package;
the fraction of the initial energy lost;
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I