On planet Tehar, the acceleration of gravity is the same as that on Earth but there is also a strong downward electric field with the field being uniform close to the planet's surface. A 2.16 kg ball having a charge of 4.80 uC is thrown upward at a speed of 21.0 m/s and it hits the ground after an interval of 3.63 s. What is the potential difference between the starting point and the top point of the trajectory?
The ball's acceleration downward is given by
a * (3.63/2 s) = 21.0 m/s
a = 11.57 m/s^2
Since g = 9.81 m/s^2, the downward acceleration due to electrostatic force is 1.76 m/s^2
The potential difference V due to electrostatic force at the top of the trajectory is given by
M*1.76* H = Q*V
Solve for V.
what is H?
To find the potential difference between the starting point and the top point of the trajectory, we can use the concept of electric potential energy.
First, let's find the initial kinetic energy (KEi) of the ball when it is thrown upward. The formula for kinetic energy is:
KE = (1/2)mv^2,
where m is the mass of the ball and v is its velocity.
Given:
Mass (m) = 2.16 kg
Velocity (v) = 21.0 m/s
Substituting these values into the formula, we have:
KEi = (1/2) * 2.16 kg * (21.0 m/s)^2
= 487.08 J
Now, let's find the final kinetic energy (KEf) of the ball when it hits the ground. The final kinetic energy is zero since the ball comes to rest.
So, KEf = 0 J
The difference between the initial and final kinetic energies gives us the change in kinetic energy (ΔKE) experienced by the ball:
ΔKE = KEf - KEi
= 0 J - 487.08 J
= -487.08 J
Since energy cannot be created or destroyed, the change in kinetic energy is equal to the work done on the ball by external forces, which includes both gravity and the electric field.
The work done by gravity is given by the formula:
Wg = mgh,
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
Given:
Mass (m) = 2.16 kg
Acceleration due to gravity (g) is the same as that on Earth
Height (h) = ?
To find the height, we need to use the time of flight of the projectile. The time of flight is the total time taken for the ball to reach its highest point and come back to the ground. It is given as 3.63 s.
The time of flight is twice the time taken to reach the maximum height, which is also the time taken for the ball to move upwards. So, the time taken to reach the maximum height is t/2 = 3.63 s / 2 = 1.815 s.
Using the equation of motion:
h = (1/2)gt^2
where h is the height, g is the acceleration due to gravity, and t is the time taken to reach the maximum height.
Substituting the values into the equation, we have:
h = (1/2) * g * (1.815 s)^2
Now we can substitute the value of height (h) into the formula for work done by gravity, Wg:
Wg = mgh
= 2.16 kg * g * [(1/2) * g * (1.815 s)^2]
Now let's find the work done by the electric field:
We know that work done by an electric field is given by:
We = qV,
where q is the charge of the ball and V is the potential difference between the starting point and the top point of the trajectory.
Given:
Charge (q) = 4.80 uC
Potential difference (V) = ?
Substituting the values into the formula, we have:
We = (4.80 uC) * V
Since the work done by external forces is equal to the change in kinetic energy, we have:
Wg + We = ΔKE
Substituting the respective formulas, we have:
2.16 kg * g * [(1/2) * g * (1.815 s)^2] + (4.80 uC) * V = -487.08 J
Now we can solve for the potential difference (V) between the starting point and the top point of the trajectory by rearranging the equation:
V = (-487.08 J - 2.16 kg * g * [(1/2) * g * (1.815 s)^2]) / (4.80 uC)
To find the value of g, the acceleration due to gravity on planet Tehar, we need to know its value or have additional information about the planet. Without that information, we cannot provide a specific answer to the potential difference between the starting point and the top point of the trajectory on planet Tehar.