Down an Incline with Spring: Finding Distance and Maximum Speed

[Albeaver89]

Homework Statement

A 3.5 kg block starts at rest and slides a distance d down a frictionless 28.0° incline, where it runs into a spring. The block slides an additional 21.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 435 N/m.
A)What is the value of d?
B)What is the distance between the point of first contact and the point where the block's speed is greatest?

Homework Equations

A)Conservation of Energy: Ei=Ef ⇔ KEi+PEi=KEf+PEf ⇔ 0+mgh=1/2kx^2
h=(Δx+d)sinθ (Δx=the distance the spring was compressed)
Δxsinθ+dsinθ=1/2kx^2 solve for d
d=(.5kx^2)/(gmsinθ)-Δx
d=.385 m or 38.5 cm

B) So the first contact is when it touches the the spring, correct? Then it's greatest speed is the speed right after it leaves the spring going up the ramp? That would mean that I need the speed (v) of the block as it hits spring equation: KEi+PEi=KEf+PEf ⇔ m*g*(d*sinθ)=1/2*k*x^2+1/2*m*(vf)^2 where vf= the speed it hits the spring.
v=√(-1/m)*√(k*x^2-2*d*g*m*sinθ)
V=1.39 m/s when it hits the spring

KEi+PEi=KEf+PEf ⇔ 1/2m(vi)^2+mgh=1/2m(vf)^2+1/2kΔx^2 where Δx= the distance the spring is compressed.

Does that seem correct?

The Attempt at a Solution

 

[SammyS]

Homework Statement

A 3.5 kg block starts at rest and slides a distance d down a frictionless 28.0° incline, where it runs into a spring. The block slides an additional 21.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 435 N/m.
A)What is the value of d?
B)What is the distance between the point of first contact and the point where the block's speed is greatest?

Homework Equations

A)Conservation of Energy: Ei=Ef ⇔ KEi+PEi=KEf+PEf ⇔ 0+mgh=1/2kx^2
h=(Δx+d)sinθ (Δx=the distance the spring was compressed)
Δxsinθ+dsinθ=1/2kx^2 solve for d
d=(.5kx^2)/(gmsinθ)-Δx
d=.385 m or 38.5 cm

It seems to me, that Δx is 21.0 cm.

Is that what you used?

The potential energy of the spring at max compression is .5k(Δx)².

B) So the first contact is when it touches the the spring, correct? Then it's greatest speed is the speed right after it leaves the spring going up the ramp? That would mean that I need the speed (v) of the block as it hits spring equation: KEi+PEi=KEf+PEf ⇔ m*g*(d*sinθ)=1/2*k*x^2+1/2*m*(vf)^2 where vf= the speed it hits the spring.
v=√(-1/m)*√(k*x^2-2*d*g*m*sinθ)
V=1.39 m/s when it hits the spring

KEi+PEi=KEf+PEf ⇔ 1/2m(vi)^2+mgh=1/2m(vf)^2+1/2kΔx^2 where Δx= the distance the spring is compressed.

Does that seem correct?

The Attempt at a Solution