I posted this about 45 min ago, then realized I made a typo and excluded nearly
ID: 1310873 • Letter: I
Question
I posted this about 45 min ago, then realized I made a typo and excluded nearly a whole line of the question. I tried to update it and correct my mistake, but Chegg wouldn't let me. Whenever I clicked the "Update" button, I got a broken link and nothing happened. So, I am reposting this question now that it is asked correctly.
A 2.50-kg block slides on a horizontal surface with coefficient of kinetic friction 0.310. It is moving at 5.45 m/s when it encounters a spring of force constant 50.0 N/m. The object comes to rest momentarily when it has compressed the spring a distance d, at which point it turns around and goes back in the direction it came from, and comes to rest at a distance D beyond the point at which it made contact with the spring.
(a) Find the distance of compression, d.
(b) Find the speed of the block on the return, at the point where it separates from the spring.
(c) Find the distance D beyond the spring where it comes to rest.
Explanation / Answer
given mass of block=2.50kg
coefficeint of friction=0.31
therefore frictional force acting on body f=0.31*2.5=0.775N
speed of block at the time of encounter=5.45m/sec
using conservation of energy concept,
kinetic enrgy of block=frictional energy+spring energy
1/2*2.5*(5.45)^2=0.775*x+1/2*50*x^2
37.13=0.775*x+25*x^2
a) therefore x=1.203m
therefore distance of compression d=1.203m
b)
1/2*k*d^2=0.775*d+1/2*m*v^2
1/2*50*(1.203)^2=0.775*1.203+1/2*2.5*v^2
36.18=0.9323+1.25*v^2
v^2=28.198
velocity v=5.31m
c)1/2*k*d^2=0.775*(d+D)
1/2*50*(1.203)^2=0.775*(1.203+D)
D=45.48m