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:
Hi there!
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:
The distance travel by block before coming to rest is 0.122 m
Explanation:
Given:
Mass of block kg
Initial speed of block
Final speed of block
Coefficient of kinetic friction
Ramp inclined at angle 28.4°
Using conservation of energy,
Work done by frictional force is equal to change in energy,
Where
m
Therefore, the distance travel by block before coming to rest is 0.122 m
(a) 12.0 V
In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, , the charged stored on the capacitor at the end of the process is
When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship
and since neither Q nor C change, the voltage V remains the same, 12.0 V.
(b) (i) 24.0 V
In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:
where:
is the permittivity of free space
is the relative permittivity of the material inside the capacitor
A is the area of the plates
d is the separation between the plates
As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: . The new voltage across the plate is given by
and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is
So, the new voltage is
(c) (ii) 3.0 V
The area of each plate of the capacitor is given by:
where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be
While the separation between the plate was unchanged (d); so, the new capacitance will be
So, the capacitance has increased by a factor 4; therefore, the new voltage is
which means
Answer:
The energy of the system is 15 J.
Explanation:
Given that,
Energy E = 2.5 J
Amplitude = 10 cm
We need to calculate the spring constant
Using formula of mechanical energy of the system
Put the value into the formula
If the block is replaced by a block with twice the mass of the original block
Amplitude = 6 cm
We need to calculate the energy
Using formula of mechanical energy
Put the value into the formula
Hence, The energy of the system is 15 J.