physics question?

Question

A 20 kg mass, released from rest, slides 6 meters down a frictionless plane inclined at an angle of 30 degrees with the horizontal and strikes a spring of spring constant k =200 newtons/meter as shown in the diagram above. Assume that the spring is ideal, that the mass of the soring is negligible, and the mechanical energy is conserved.
a) Determine the speed of the block just before it hits the spring.
b) Determine the distance the spring has been compressed when the block comes to rest?
c) Is the speed of the block a maximum at the instant the block strikes the spring?
d) if there is friction and a coefficient of .2 what is the speed of the block just prior to hitting the spring?
e) Imagine that there is no friction in the region where the spring is. What would be the distance that the block would travel up the incline after it rebounded from the spring?

Answer 1

No diagram. I will assume the spring is positioned on the incline so the the block continues to have gravity pulling as the spring is compressed

using energy
m*g*(6+x)*sin(30)=.5*k*x^2
solve for x
b) x=2.97 m

a) m*g*6*sin(30)=.5*m*v^2
solve for v
a) v=sqrt(9.81*6)
v=7.67 m/s

c) For 0 m*g*(6+x)*sin(30)=.5*k*x^2+.5*m*v^2
rearrange and simplify
v=sqrt(9.81(6+x)-10*x^2)

take dv/dx and set = 0

9.81=20*x
x=9.81/20
x=0.49
check the speed when x=0.49


v=7.827, that's the max speed

d)
m*g*6*sin(30)-m*g*cos(30)*0.2*6
=.5*m*v^2

solve for v
v=6.202 m/s

e) since there is no friction durin the compression and release of the spring, there is no energy lost.
the height, then is
m*g*h*sin(30)+m*g*cos(30)*0.2*h=
.5*m*v^2

solve for h with v=6.202

h=2.91 m

j