A 500g ball is dropped into a well that is 10m deep. A)What is the ball's potential energy before it is dropped? Take the bottom of the well to be at zero height. I solved by using PE=mgh and came up with 49J. B) What are the kinetic energy and potential energy when it has fallen 7m? How fast is it traveling at this point? I used PE=mgh but used 3m for h and came up with 14.70J. Then solved KE=1/2MV^2 and used 7m for h and came up with 34.28J. I know these two have to equal the starting PE due to law of conservation of energy which they come pretty close. Did I do this right or did I use the wrong numbers in h?..Then for the last part I used V=sqrt2gh and came up with 11.71 m/s. I appreciate any assistance or pointers on this.

Question

A 500g ball is dropped into a well that is 10m deep. A)What is the ball's potential energy before it is dropped? Take the bottom of the well to be at zero height. I solved by using PE=mgh and came up with 49J. B) What are the kinetic energy and potential energy when it has fallen 7m? How fast is it traveling at this point? I used PE=mgh but used 3m for h and came up with 14.70J. Then solved KE=1/2MV^2 and used 7m for h and came up with 34.28J. I know these two have to equal the starting PE due to law of conservation of energy which they come pretty close. Did I do this right or did I use the wrong numbers in h?..Then for the last part I used V=sqrt2gh and came up with 11.71 m/s. I appreciate any assistance or pointers on this.

Answer 1

A) To find the potential energy (PE) of an object, you correctly used the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

Given that the mass of the ball is 500g (which is equivalent to 0.5kg) and the height of the well is 10m, you substituted these values into the equation as follows:

PE = (0.5kg) * (9.8m/s^2) * (10m) = 49J

So, your calculation of the ball's potential energy before it is dropped into the well is correct.

B) Now, let's consider the situation when the ball has fallen 7m from the top of the well.

To find the potential energy at this point, you need to use the new height (h) of 7m and the same mass. Applying the equation gives:

PE = (0.5kg) * (9.8m/s^2) * (7m) = 34.3J (rounded to two decimal places)

Therefore, the potential energy of the ball when it has fallen 7m is approximately 34.3J.

To find the kinetic energy (KE) at this point, you can use the law of conservation of energy. Since the ball's starting potential energy is 49J, the total mechanical energy (PE + KE) should remain constant.

Total mechanical energy = PE + KE

49J = 34.3J + KE

KE = 49J - 34.3J = 14.7J

So, the kinetic energy of the ball when it has fallen 7m is approximately 14.7J.

To determine the velocity (V) of the ball at this point, you correctly used the equation KE = (1/2)mv^2, where m is the mass and V is the velocity.

Substituting the known values (mass = 0.5kg and KE = 14.7J) into the equation, you can solve for V:

14.7J = (1/2) * (0.5kg) * V^2

V^2 = (14.7J * 2) / (0.5kg)

V^2 = 58.8 m^2/s^2

V ≈ √58.8 ≈ 7.67 m/s (rounded to two decimal places)

Therefore, you have correctly determined the ball's kinetic energy, potential energy, and velocity when it has fallen 7m into the well. Your calculations and use of the formulas appear to be correct.