A block of mass m = 3.20kg slides from rest a distance d down a frictionless incline at angle θ = 30.0° where it runs into a spring of spring constant 425 N/m. When the block momentarily stops, it has compressed the spring by 22.0 cm.

Question

(a) What is the distance d?
I found this to be 0.436 meters.
(b) What is the distance between the point of first contact and the point where the block's speed is greatest?
I cannot figure this one out. I thought it would just be zero, but that answer isn't correct..

Answer 1

The block's speed is greatest when the acceleration (and net acting force down the incline) are zero.

Let the displacement from the point of first contact be x.

Solve
kx = M g (d + x)sin30

Answer 2

Okay, here is my work:
kx=Mg(d+x)sin30
425x=(3.2)(9.8)(.436+x)sin30
425x=15.68(.436+x)
425x=6.83648+15.68x
409.32x=6.83648
x=59.8729 m
And then, converted to cm (as required) would be 5987.29. Is this correct?

Answer 3

your explanation is quite inadequate and lacking, "drwls". sigh.