What is the direction of the frictional force?
Answer:
Direction of friction is always opposite to relative motion
Explanation:
Friction always resist the relative motion of two surface so it is always opposite to the direction of relative velocity
What is the direction of the normal force?
Answer:
It is perpendicular to the contact plane
Explanation:
Normal force is the perpendicular force which is always at 90 degree with the contact surface
How does the frictional force depend on the normal force?
Answer:
Explanation:
Friction force is directly proportional to the normal force so we can say it is given as
Let the x-axis be parallel to the incline and the y-axis rise from the incline at a right angle. By Newton's second law, if there is no acceleration in the y-direction (perpendicular to the plane), what must be the magnitude of the normal force?
Answer:
Explanation:
Normal force is equal to the component of the weight opposite to the direction of normal force
Use Newton's second law and the x-components of the forces to find the acceleration. m/s2
Answer:
Explanation:
As we know that
Answer:
20 [N], in the opposite direction of the first force.
Explanation:
We know that newton's second law stipulates that the sum of forces on a body must be equal to the product of mass by acceleration.
The negative sign means that the other force acting on the body must be in the opposite direction to the force of 30 [N]
(a) On the coil: 20 V, on the resistor: 0 V
The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:
The potential difference across the inductance is given by
(1)
where
is the time constant of the circuit
At time t=0,
So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:
(b) On the coil: 0 V, on the resistor: 20 V
Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for
Since is at the order of less than milliseconds.
Using eq.(1), we see that for , the exponential becomes zero, and therefore the potential difference across the coil is zero:
Therefore, the potential difference across the resistor will be
(c) Yes
The two voltages will be equal when:
(2)
Reminding also that the sum of the two voltages must be equal to the voltage of the battery:
And rewriting this equation,
Substituting into (2) we find
So, the two voltages will be equal when they are both equal to 10 V.
(d) at
We said that the two voltages will be equal when
Using eq.(1), and this last equation, this means
And solving the equation for t, we find the time t at which the two voltages are equal:
(e-a) -19.2 V on the coil, 19.2 V on the resistor
Here we have that the current in the circuit is
The problem says this current is stable: this means that we are in a situation in which , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law
Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.
The voltage on the coil is given by
(1)
which means that it is maximum at the moment when the battery is disconnected, when t=0:
And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.
(e-b) 0 V on both
After several seconds,
If we use this approximation into the formula
(1)
We find that
And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.