Q9: Suppose that the metal cylinder in the last problem has a mass of 0.10 kg and the coefficient of static friction between the
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Answer:
a) The gravitational potential energy before the cart rolls down the incline is 24.6 J.
b) The magnitude of the force that causes the cart to roll down is 5.47 N.
c) The acceleration of the cart is 2.38 m/s²
d) It takes the cart 1.94 s to reach the bottom of the incline.
e) The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.
f) The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.
g) The work done by the gravitational force is 24.6 J.
Explanation:
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a) The gravitational potential energy is calculated using the following equation:
EP = m · g · h
Where:
EP = gravitational potential energy.
m = mass of the object.
g = acceleration due to gravity.
h = height at which the object is located.
The height of the inclined plane can be calculated using trigonomoetry:
sin 14.0° = height / lenght
sin 14.0° = height / 4.50 m
4.50 m · sin 14.0° = height
height = 1.09 m
Then, the gravitational potential energy will be:
EP = m · g · h
EP = 2.30 kg · 9.81 m/s² · 1.09 m = 24.6 J
The gravitational potential energy before the cart rolls down the incline is 24.6 J.
b) Please, see the attached figure for a graphical description of the problem and the forces acting on the cart. The force that causes the cart to accelerate down the incline is the horizontal component of the weight (Fwx in the figure). The magnitude of this force can be obtained using trigonometry:
sin 14° = Fwx / Fw
The weight of the cart (Fw) is calculated as follows:
Fw = m · g
Fw = 2.30 kg · 9.81 m/s²
Fw = 22.6 N
Then, the x-component of the weight will be:
FW · sin 14° = Fwx
22.6 N · sin 14° = Fwx
Fwx = 5.47 N
The magnitude of the force that causes the cart to roll down is 5.47 N.
c)Using the equation of Fwx we can calculate the acceleration of the cart:
Fwx = m · a
Where "m" is the mass of the cart and "a" is the acceleration.
Fwx / m = a
5.47 N / 2.30 kg = a
a = 2.38 m/s²
The acceleration of the cart is 2.38 m/s²
d) To calculate the time it takes the cart to reach the bottom of the incline, let´s use the equation of position of the cart:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the cart at time t.
x0 = initial position.
v0 = initial velocity.
a = acceleration.
t = time.
Considering the initial position as the point at which the cart starts rolling (x0 = 0) and knowing that the cart starts from rest (v0 = 0), let´s find the time it takes the cart to travel the 4.50 m of the inclined plane:
x = 1/2 · a · t²
4.50 m = 1/2 · 2.38 m/s² · t²
2 · 4.50 m / 2.38 m/s² = t²
t = 1.94 s
It takes the cart 1.94 s to reach the bottom of the incline.
e) The velocity of the cart at the bottom of the inclined plane can be obtained using the following equation:
v = v0 + a · t
v = 0 m/s + 2.38 m/s² · 1.94 s
v = 4.62 m/s
The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.
f) The kinetic energy can be calculated using the following equation:
KE = 1/2 · m · v²
Where:
KE = kinetic energy.
m = mass of the cart.
v = velocity of the cart.
KE = 1/2 · 2.30 kg · (4.62 m/s)²
KE = 24.6 J
The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.
The gain of kinetic energy is equal to the loss of gravitational potential energy.
g) The work done by the gravitational force can be calculated using the work-energy theorem: the work done by the gravitational force is equal to the negative change in the gravitational potential energy:
W = -ΔPE
W = -(final potential energy - initial potential energy)
W = -(0 - 24.6 J)
W = 24.6 J
This can also be calculated using the definition of work:
W = Fw · d
Where "d" is the distance traveled in the direction of the force, that is the height of the inclined plane:
W = 22.6 N · 1.09 m = 24.6 J.
The work done by the gravitational force is 24.6 J.
Answer:
<em>radius of the loop = 7.9 mm</em>
<em>number of turns N ≅ 399 turns</em>
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ = = 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop =
0.005 = 2 x 3.142 x r
r = 0.005/6.284 = = 0.0079 m =<em> 7.9 mm</em>