AT&T; 10:15 PM The Expert TA I Human-like Grading, Automated! usi38ok.theexpertt

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AT&T; 10:15 PM The Expert TA I Human-like Grading, Automated! usi38ok.theexpertta.com Homework 6 Begin Date: 3/9/2018 12:00:00 AMDue Date: 3/28/2018 11:59:00 PM End Date: 3/28/2018 11:59:00 PM (B%) ProblemN: Abhok ofmass 5.3kgus aiting on a fri tomless tansp with a angle ofthe ramp with respect ?, the bowetalsle. 33% Part (a) The block, starting from rest, slides down the ramp a distance 45 cm before hitting the spring. How far, in centimeters, is the spring compressed as the block comes to momentary rest? OD pm Raduni Submission History 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 .. 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? All content 298 Expert T

Explanation / Answer

a) Apply conservation of energy

final elastic potential energy = loss of gravitational potential energy

(1/2)*k*x^2 = m*g*(d+x)*sin(14)

(1/2)*580*x^2 = 5.3*9.8*(0.45 + x)*sin(14)

==> x = 0.163 m <<<<<<<<<<<-----------------Answer

b) v = sqrt(2*g*d*sin(14))

= sqrt(2*9.8*0.45*sin(14))

= 1.46 m/s <<<<<<<<<<<-----------------Answer

c) change in potential energy = -m*g*(d+x)

= -5.3*9.8*(0.45 + 0.163)

= -31.8 J <<<<<<<<<<<-----------------Answer

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