PE =KE =PE (of the spring)
m•g•h = m•v^2/2=k•x^2/2,
h=d•sinα
m•g• d•sinα = k•x^2/2
d=k•x^2/2• m•g•sinα =
=421•(0.24)^2/2•4.6•9.8•sin31o=0.522 m.
The point where the speed is the greatest
is the same point where it was the first
contact with the spring.
A 4.6 kg block starts at rest and slides a distance d down a frictionless 31.0° incline, where it runs into a spring. The block slides an additional 24.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 421 N/m. What is the value of d?
What is the distance between the point of first contact and the point where the block's speed is greatest?
2 answers
don't follow^ webassign says its wrong so Elena needs to square the up cuz she trippin