Class Management I Help HW 5 Begin Date: 10/1/2017 12:00:00 AM-Due Date: 10/10/2
ID: 1777324 • Letter: C
Question
Class Management I Help HW 5 Begin Date: 10/1/2017 12:00:00 AM-Due Date: 10/10/2017 11:59:00 PM End Date: 12/16/2017 12:00:00 AM (696) 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 table, compressing the spring a distance of d-o 18 meters. The spring constant of the spring is 890 Nm. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring Randomized Variables m 0.11 kg k=890 N/m d= 0.18 m 33% Part (a) The ball is then released what is its speed v, in meters per second, Just after the ball leaves the spring? Grade Summary Potential 100% Submissions Attempts remaining 2 detailed view 0% sino cos) cotan asin acos0 atan) acotan sinh0 cosh)tnh cotanh) per attempt) DegreesORadians Sulbemit Hint 1 give I give upt Submit Hints:1% deduction per hint Hmts remaining Feedback: -deduction per feedback. 33% Part (b) what is the maximum height, h, in meters, that the ball reaches above the equilibrium point? 33% 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
here,
m = 0.11 kg
d = 0.18 m
k = 890 N/m
a)
let the speed after release be v
using conservation of energy
0.5 * m * v^2 + m * g * d = 0.5 * k * d^2
0.5 * 0.11 * v^2 + 0.11 * 9.81 * 0.18 = 0.5 * 890 * 0.18^2
solving for v
v = 16.1 m/s
b)
let the maximum height be h
using conservation of energy
0.5 * m * v^2 = m * g * h
0.5 * 16.1^2 = 9.81 * h
h = 13.2 m
c)
let the ball's velocity at half of maximum height be v'
using conservation of energy
0.5 * m * v^2 = m * g * h/2 + 0.5 * m* v'^2
0.5 * 16.1^2 = 9.81 * 13.2 /2 + 0.5 * v'^2
v' = 11.4 m/s