Class Management | Help Due Date: 1029 2017 11:59:00 PM Eud Date: Chapter 8- Pot
ID: 1790880 • Letter: C
Question
Class Management | Help Due Date: 1029 2017 11:59:00 PM Eud Date: Chapter 8- Potential Energy and Conservation of En Begin Date: 10 12 2017 3:00:00 PM 10/29/2017 11:59:00 PM (17%) Problem 5: A student pushes a baseball of m-0.11 kg down onto the top of a vertical spring that has its lower end fixed to a tablc, comprcssing thc spring a distancc of d 0.14 mctcrs. The spring constant of thc spring is 850 N/m. Lct tho gravitational potential cncrgy be zcro at the position of the bascball in the compressed spring. Randomized Variables m=0.11kg k 850 N/m d- 0.14 nm - 33% Part (a) 1he ball is then released. What is its speed, v, in meters per second, just after the ball leaves the spring? Grade Summary Deductioras Potential 090 10096 Submissions cotan asinacoso ataacotansinh cosh tanotnh Attempts remaining: (1 % per attempt) detailed view Dcgrccs Radians Submit Hint I give up! Hints! (1% deduction per hint. Hints remaining: Feedback: 1% dedn ner feedback. - 33% Part(b) What is the maximum height, h, in meters that the ball reaches above the equilibrium point? 3396 Part (c) What is the ball's velocity in meters per second, at half of the maximum height relative to the equilibrium point?Explanation / Answer
(A) Applying work - energy theorem,
Work done by gravity + work done by spring = change in KE
- 0.11 x 9.81 x 0.14 + (850)(0.14^2)/2 = 0.11 (v^2 - 0) / 2
v = 12.2 m/s
(B) After that ,
PEi + KEi = PEf + KEf
0 + m v0^2 /2 = m g h + 0
h = 12.2^2 / (2 x 9.81) = 7.6 m
(C) h' = h/2 = 3.78 m
12.2^2 / 2 = (9.81 x 3.78) + v^2 /2
v = 8.63 m/s