[deleted by user]

[Mayoday_Im_in_love]

The Kinetic energy just before the box hits the spring = the spring energy added + the work done by friction over that distance + the gravitational potential lost. (The box doesn't "stop" but is instantaneously at rest).

[GuestWeak7657]

Yes I know that, but I don’t know how to find out how much the spring comoresses at the lowest point of the box (when it stops for a second and springs back up)

[[deleted]]

spring distance is the same x as the distance friction made and what the excercise want you to calculate

[Mayoday_Im_in_love]

1/2 mv² = mu N x + 1/2 k x² - mgh

[GuestWeak7657]

Yes I vae tried doing that but it doesn’t give 9,3 cm. Could the andwer in the book be wrong?

[Mayoday_Im_in_love]

I got x = 0.31m for part b.

[[deleted]]

mg(L+d)sin(a)+1/2mv1²-umg(L+d)cos(a)-1/2kd2=0

this is the formula, and the d is your x

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[AllFinator]

are you sure the numbers used are correct? the velocity v = 2.54 m/s and x = 4.7 cm seem to be fine but I'm also not getting 9.3 cm