Class Management Help IIW 10 -Part B Begin Date: 3/15/2018 12:01:00 AM - Due Dat
ID: 2030652 • Letter: C
Question
Class Management Help IIW 10 -Part B Begin Date: 3/15/2018 12:01:00 AM - Due Date: 3/24/2018 11:59:00 PM End Date: 3/31/2018 11:59:00 PM (2006) Problem 4: A block of mass 3.6 kg is sitting on a frictionless ramp with a spring at the bottom that has a spring constant of 490 N/m (refer to the figure). The angle of the ramp with respect to the horizontal is 23o. Otheexpertta.com 33% Part (a) The block, starting from rest, slides down the ramp a distance 77 cm before hitting the spring. How far, in centimeters s the spring compressed as the block comes to momentary rest? Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 10 (0per attempt) Sin cos tanO Tt( acoso acotan sinh) cosh tanh) cotanho cotan) asin() detailed view 0 END Degrees Radians NO Submit Hint I give up! llints: 2 deduction per hint. Ilints remaining: 3 Feedback:-%" deduction per feedback. 33% Part (b) After the block comes to rest, the spring pushes the block back up the ramp. How fast in meters per second, is the block moving right after it comes off the spring? 33% Part c) what is the change of the gravitational potential energy in joules, between the original position of the block at the top of the ramp and the position of the block when the spring is fully compressed?Explanation / Answer
Part (a) -
Downward acceleration of the block along the ramp, a = g*sin23 = 9.81*sin23 = 3.83 m/s^2
so, velocity of the block after travelling a distance of s = 77 cm = 0.77 m -
v = sqrt[2*a*s] = sqrt[2*3.83*0.77] = 2.43 m/s
suppose x is the compression in the spring.
apply conservation of energy -
(1/2)*k*x^2 = (1/2)*m*v^2
=> k*x^2 = m*v^2
=> 490*x^2 = 3.6*2.43^2
=> x = sqrt[(3.6*2.43^2) / 490] = 0.208 m = 20.8 cm
Part (b) -
The velocity of the block after it comes off the spring will be the same as its initial velocity (at the time of striking the spring) = 2.43 m/s
Part (c) -
Initial potential energy of the block, Pi = m*g*h = 3.6*9.81*[(0.77+0.208)*sin23) = 13.50 J
Final potential energy,Pf = 0
So, change in the potential energy = Pi - Pf = 13.50 - 0 = 13.50 J