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. correct

b. correct
last part: correct.

Answer 2

A) To determine the potential energy (PE) of the ball before it is dropped, you correctly used the formula PE = mgh. Here, m represents the mass of the ball (500g or 0.5kg), g represents the acceleration due to gravity (approximately 9.8 m/s^2), and h represents the height of the well (10m).

So, substituting these values into the formula, we get:
PE = (0.5kg)(9.8 m/s^2)(10m) = 49 Joules (J)

Therefore, the ball's potential energy before it is dropped is indeed 49J.

B) Now, let's calculate the kinetic energy (KE) and potential energy (PE) when the ball has fallen 7m into the well.

To find PE, we still use the formula PE = mgh, but now the height (h) is 7m (the remaining distance to the bottom of the well).

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

The difference in potential energy between the starting point and this point is:
PE difference = PE_initial - PE_final = 49J - 34.3J = 14.7J

So, you correctly determined the potential energy at this point, which is 14.7J.

To find the kinetic energy (KE) at this point, we use the conservation of energy principle, which states that the total mechanical energy remains constant throughout the ball's motion.

Total mechanical energy (E) = PE + KE

Initially, the total mechanical energy (E_initial) is equal to the potential energy (PE_initial) since the ball is at rest. So, E_initial = PE_initial = 49J.

At the point where the ball has fallen 7m, the total mechanical energy (E_final) will still be equal to the initial total mechanical energy:
E_final = E_initial = 49J

Therefore, the kinetic energy (KE_final) at this point is:
KE_final = E_final - PE_final = 49J - 34.3J = 14.7J

Hence, the kinetic energy at this point is also 14.7J, which matches the potential energy.

Finally, to determine the speed (V) of the ball at this point, we use the equation: KE = 1/2mv^2, where m is the mass (0.5kg) and v is the velocity of the ball.

Rearranging the equation, we have:
v = sqrt(2KE/m)

Substituting the values, we get:
v = sqrt[(2*14.7J)/(0.5kg)] ≈ 11.7 m/s

So, you correctly calculated the speed of the ball at this point, which is approximately 11.7 m/s.

In conclusion, you have solved the problem correctly by using the formulas and the correct values. Well done!