A 5.0 ball is initially sitting at rest on top of a 41 m high hill. The ball is pushed with a force of 25 N for 3.0 s toward the edge of a hill. Upon reaching the edge, the ball is allowed to roll down the hill where it reaches the flat ground. To the nearest mile per hour, how fast is the ball traveling when it reaches the flat ground at the bottom of the hill? neglect all types of friction
To find the velocity of the ball when it reaches the flat ground at the bottom of the hill, we can use the principle of conservation of energy.
Step 1: Calculate the potential energy of the ball at the top of the hill.
The potential energy (PE) of an object is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. Since the ball is initially at rest, its initial kinetic energy is zero.
Given:
Mass of the ball (m) = 5.0 kg
Height of the hill (h) = 41 m
Acceleration due to gravity (g) ≈ 9.8 m/s² (neglecting decimals for simplicity)
Potential Energy (PE) = mgh
PE = 5.0 kg * 9.8 m/s² * 41 m
Step 2: Calculate the final kinetic energy of the ball at the flat ground.
When the ball reaches the flat ground, it no longer has any potential energy but only kinetic energy. The final kinetic energy (KE) is given by the formula KE = 1/2 * mv², where v is the velocity of the ball.
Since we want to find the velocity in mph (miles per hour), we need to convert the units.
Given:
1 mile = 1609.34 meters
1 hour = 3600 seconds
Use these conversion factors to convert the final velocity to mph.
Step 3: Equate the potential energy to the kinetic energy.
Since energy is conserved, equating the potential energy (PE) to the kinetic energy (KE):
PE = KE
mgh = 1/2 * mv²
Step 4: Solve for the velocity.
Divide both sides of the equation by (1/2 * m) to isolate v²:
v² = 2gh
Now, substitute the values and calculate:
v² = 2 * 9.8 m/s² * 41 m
v² ≈ 803.6 m²/s²
Convert meters per second to miles per hour:
v (in mph) = (v (in m/s) * 3600 seconds) / 1609.34 meters
v (in mph) ≈ (√(803.6 m²/s²) * 3600 seconds) / 1609.34 meters
v (in mph) ≈ (√(803.6) * 3600) / 1609.34
v (in mph) ≈ (28.37 * 3600) / 1609.34
v (in mph) ≈ 63.78 mph
Therefore, to the nearest mile per hour, the speed of the ball when it reaches the flat ground at the bottom of the hill is approximately 63.78 mph.
To calculate the speed of the ball when it reaches the flat ground at the bottom of the hill, we need to consider the principles of energy conservation and kinematics.
First, let's calculate the potential energy of the ball at the top of the hill using the formula: potential energy (PE) = mass (m) x acceleration due to gravity (g) x height (h).
PE = 5.0 kg x 9.8 m/s² x 41 m
PE = 2009 J
Next, let's consider the work done on the ball by the applied force. Since the ball is being pushed horizontally, the work done on the ball will be equal to the force applied (F) multiplied by the distance (d) the ball moves.
Work (W) = Force (F) x Distance (d)
W = 25 N x d
To find the distance, we need to calculate the average velocity of the ball during the 3.0 s pushing duration. The formula for average velocity (v) is:
v = d / t
where d is the distance and t is the time.
Using the formula,
25 N x d = 5.0 kg x (d/3.0 s)
Simplifying,
25 N x d = 1.67 kg m/s x d
Now we can solve for the distance d:
25 N x d = 1.67 kg m/s x d
25 N = 1.67 kg m/s
d = 1.67 kg m/s / 25 N
d = 0.0668 m
Now that we know the distance, we can calculate the work done on the ball:
W = 25 N x 0.0668 m
W = 1.67 J
Since the work done on the ball is equal to the change in potential energy, we can write:
W = ΔPE
1.67 J = ΔPE
To find the final kinetic energy (KE) of the ball when it reaches the bottom of the hill, we can equate the change in potential energy to the change in kinetic energy using the formula:
ΔPE = ΔKE
1.67 J = KE - 0 J (since the ball is initially at rest)
Therefore, the final kinetic energy of the ball is:
KE = 1.67 J
Finally, let's calculate the speed of the ball using the formula for kinetic energy:
KE = 1/2 x mass x velocity²
Rearranging the formula and plugging in the values:
1.67 J = (1/2) x 5.0 kg x v²
v² = (2 x 1.67 J) / 5.0 kg
v² = 0.67 J / 5.0 kg
v² = 0.134 J/kg
v = √0.134 J/kg
To convert the speed of the ball from meters per second to miles per hour, we can multiply by a conversion factor:
v (mph) = (v (m/s) x 2.2369)
Therefore, the speed of the ball when it reaches the flat ground at the bottom of the hill is approximately:
v (mph) ≈ (√0.134 J/kg x 2.2369) mph
By solving this equation, we can obtain the final answer in miles per hour.