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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?
What is the distance between the point of first contact and the point where the block's speed is greatest?
Answers
Answered by
Elena
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.
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.
Answered by
Elena you suck
don't follow^ webassign says its wrong so Elena needs to square the up cuz she trippin
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