a 5.4 kg ball slides along a friction-less, horizontal surface at 6.2 m/s. The ball then encounters a ramp which slopes upward. The ball slides up the ramp, slowing down until it stops and slides back down. what is the kinetic energy of the ball as it slides along the friction- less, horizontal surface? What is the maximum height reached by the ball as it slides up the slope?
ke = (1/2)(5.4) (6.2)^2
stops so all ke converted to U
U = m g h = ke from first part
5.4 (9.81) h = answer from part 1
To find the kinetic energy of the ball as it slides along the friction-less, horizontal surface, we can use the formula for kinetic energy:
Kinetic Energy (KE) = (1/2) * mass * velocity^2
Given that the mass of the ball (m) is 5.4 kg and its velocity (v) is 6.2 m/s, we can substitute these values into the formula:
KE = (1/2) * 5.4 kg * (6.2 m/s)^2
= (1/2) * 5.4 kg * 38.44 m^2/s^2
= 103.644 J
Therefore, the kinetic energy of the ball as it slides along the friction-less, horizontal surface is 103.644 Joules.
To determine the maximum height reached by the ball as it slides up the slope, we need to consider the conservation of energy. The ball loses kinetic energy as it slows down while climbing the ramp, which is converted into potential energy as it gains height.
The total mechanical energy (E) of the ball, which is the sum of kinetic and potential energy, remains constant:
E = KE + PE = constant
Since the ball starts from a horizontal surface and comes to a stop at the highest point on the ramp, its final kinetic energy at the top is zero. Therefore, the initial kinetic energy (KE) of the ball is equal to the potential energy (PE) it gains at the maximum height.
PE = m * g * h
Where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Therefore, we can rewrite the equation as:
KE = PE
m * (v^2)/2 = m * g * h
Substituting the given values:
5.4 kg * (6.2 m/s)^2 / 2 = 5.4 kg * 9.8 m/s^2 * h
Simplifying the equation:
19.62 J = 52.92 kg*m^2/s^2 * h
Dividing both sides by 52.92 kg*m^2/s^2:
h = 19.62 J / (52.92 kg*m^2/s^2)
Calculating the value:
h ≈ 0.370 m
Therefore, the maximum height reached by the ball as it slides up the slope is approximately 0.370 meters.