A 40 kg ball is pukled to one side until it is 1.5 above itd lowest point. What will be its velocity as it passes through its lowest point
the potential energy at the high point is equal to the kinetic energy at the low point
m g h = 1/2 m v^2 ... v = √(2 g h)
the mass is in kilograms , so the height is probably in meters
v = √(2 * 9.8 * 1.5) m/s
Well, if a 40 kg ball is being "pukled" to one side, I'm assuming you meant pulled, not puked. I hope nobody is puking on the poor ball!
Now, as for its velocity as it passes through its lowest point, let's find out. Since only the height difference is given, we need more information to calculate the velocity. We would need to know the initial velocity or force applied to the ball. Otherwise, my crystal ball says, "Sorry, I can't predict that!"
To determine the velocity of the ball as it passes through its lowest point, we can use the principle of conservation of mechanical energy.
The mechanical energy of an object consists of kinetic energy (KE) and potential energy (PE). At the lowest point of the ball's motion, all of its potential energy is converted into kinetic energy.
Given:
Mass of the ball (m) = 40 kg
Height above the lowest point (h) = 1.5 m
To calculate the velocity, we need to find the potential energy (PE) at the starting point and equate it to the kinetic energy (KE) at the lowest point.
1. Calculate the potential energy at the starting point:
PE = m * g * h
where m = mass of the ball, g = acceleration due to gravity (approximately 9.8 m/s^2), and h = height above the lowest point.
PE = 40 kg * 9.8 m/s^2 * 1.5 m
PE = 588 Joules
2. The potential energy at the starting point is equal to the kinetic energy at the lowest point:
KE = 1/2 * m * v^2
where v = velocity of the ball at the lowest point.
KE = 588 Joules
Now, we can solve for the velocity (v):
1/2 * 40 kg * v^2 = 588 Joules
20 kg * v^2 = 588 Joules
v^2 = 588 Joules / 20 kg
v^2 = 29.4 m^2/s^2
Taking the square root of both sides:
v = √(29.4 m^2/s^2)
v ≈ 5.42 m/s
Therefore, the velocity of the ball as it passes through its lowest point is approximately 5.42 m/s.
To determine the velocity of the ball as it passes through its lowest point, we can use the principle of conservation of energy. This principle states that the total mechanical energy of a system remains constant if no external work is done on the system.
In this case, the ball is being pulled to one side and lifted to a height of 1.5 meters above its lowest point. As it passes through its lowest point, all of its potential energy will be converted into kinetic energy. Therefore, we can equate the potential energy of the ball at its highest point to the kinetic energy of the ball at its lowest point.
The potential energy (PE) of an object is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height relative to a reference point.
In this case, the mass of the ball (m) is 40 kg, and the height (h) is 1.5 meters. Substituting these values into the equation, we get:
PE = (40 kg) * (9.8 m/s^2) * (1.5 m)
= 588 Joules
The kinetic energy (KE) of an object is given by the equation KE = (1/2)mv^2, where v is the velocity of the object.
Since all the potential energy is converted into kinetic energy, we can set PE equal to KE:
PE = KE
588 Joules = (1/2)(40 kg)v^2
Simplifying the equation, we can solve for v:
v^2 = (2 * 588 Joules) / (40 kg)
v^2 = 11.76 m^2/s^2
v ≈ √11.76 ≈ 3.43 m/s
Therefore, the velocity of the ball as it passes through its lowest point is approximately 3.43 m/s.
Is this a SCIENCE question?
Is the ball PULLED?
1.5 WHAT above ITS lowest point?