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.

(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..

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

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

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?

To solve this problem, we can use the conservation of mechanical energy principle, which states that the total mechanical energy of a system remains constant if no external forces act on it.

(a) To find the distance d, we can equate the potential energy lost by the block with the potential energy gained by the compressed spring. The potential energy lost is given by the change in gravitational potential energy, and the potential energy gained is given by the spring potential energy.

1. Calculate the potential energy lost:
The initial potential energy of the block is given by: U1 = m * g * d * sin(θ), where g is the acceleration due to gravity (9.8 m/s^2) and θ is the angle of the incline (30°).

2. Calculate the potential energy gained:
The potential energy gained by the compressed spring is given by: U2 = (1/2) * k * x^2, where k is the spring constant (425 N/m) and x is the compression of the spring (22.0 cm = 0.22 m).

Since the total mechanical energy is conserved, we have:
U1 = U2

Substituting the values and solving for d:
m * g * d * sin(θ) = (1/2) * k * x^2
3.20 * 9.8 * d * sin(30°) = (1/2) * 425 * (0.22)^2
98.88 * d = 10.97
d = 10.97 / 98.88
d ≈ 0.1111 m

So, the distance d is approximately 0.1111 m or 11.11 cm.

(b) To find the distance between the point of first contact and the point where the block's speed is greatest, we first need to determine the maximum compression of the spring.

The maximum compression of the spring occurs when the block momentarily stops, indicating that all its initial kinetic energy has been transferred to the spring potential energy.

1. Calculate the initial kinetic energy:
The initial kinetic energy of the block is given by: K1 = (1/2) * m * v^2, where v is the initial velocity.

Since the block starts from rest, the initial kinetic energy is zero.

2. Calculate the final potential energy gained by the compressed spring:
The final potential energy gained by the compressed spring is given by: U2 = (1/2) * k * x^2, where x is the maximum compression of the spring.

Since the final kinetic energy is zero, the initial kinetic energy is transferred entirely to the spring potential energy. This gives us:
K1 = U2
(1/2) * m * v^2 = (1/2) * k * x^2

Substituting the values into the equation:
(1/2) * 3.20 * 0^2 = (1/2) * 425 * x^2
0 = 212.5 * x^2

Solving for x:
x^2 = 0
x = 0

Since x = 0, it implies that the block does not compress the spring when it reaches its maximum speed. Therefore, the distance between the point of first contact and the point where the block's speed is greatest is zero.

Please note that the answer may seem counterintuitive, but it is because the frictionless incline allows the block to reach its maximum speed even before it reaches the spring.