118.940 Home 1 Student: edayanovgmal.com Class M Chapter? Begin Date: 7/10/2018
ID: 1865455 • Letter: 1
Question
118.940 Home 1 Student: edayanovgmal.com Class M Chapter? Begin Date: 7/10/2018 1200.00 AM-Due Date: 7/17/2018 11.00.00 AM End Date: /17/2018 11.00.00 AM (10% ) Problem 8: A brick of mass m-035 kg is set against a spring with a spring constant of ky 618 N/m which has been compressed by a distance of O.1m.Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k 301 N/m. ment Status ck here for tailed vievw em Status Ctheesperita.com üe33% Parl (a) How far d2, in meters, will the second spring compress when the brick runs ?toit? ompleted @f completed | ||DR33%Parte, Completed-0081 33% Part a) How fast v.-meters per second, will the brck be moving when strikes the second spring? 33% Part (e) Now assume friction is present on the surface in between the ends of the springs ar their equilibrium lengths. mee cores et per second, of kinetic friction is -03. If the distance between the springs is x 1m. how far dg, in meters, will the second spring now compress? Completed 8 Partial 9 Completed 10 Completed sin tan 789 cotan) asin atano acotan) sinho cosh tanh cotanho °Degrees Radians Submission History MacBook Pro 2 3 5 6Explanation / Answer
(c)
the energy lost in friction ,
E= umgL= 0.5*0.35* 9.8* 1= 1.715 J
Now using conservation of energy
Initial energy in system= frictional loss+ energy left
0.5 *618*0.1^2=1.715 + 0.5*301*d^2
d= 0.0956 m
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Comment in case any doubt..goodluck
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