A 190 g block is pressed against a spring of force constant 1. 60 kn/m until the block compresses the spring 10. 0 cm. The spring rests at the bottom of a ramp inclined at 60. 0° to the horizontal. Using energy considerations, determine how far up the incline (in m) the block moves from its initial position before it stops under the following conditions.

From the law of conservation of energy, the total mechanical energy at the bottom of the incline, E equals the total mechanical energy when the mass stops, E'

So, E = E'

U + P + K = U' + P' + K' where

U = initial elastic potential energy = 1/2kx² where k = spring force constant = 1.60 kN/m = 1.60 × 10³ N/m and x = spring compression = 10 cm = 0.1 m,
P = initial gravitational potential energy = 0 J,
K = initial kinetic energy = 0,
U' = final elastice potential energy = 0,
P' = final gravitational potential energy = mgh where m = mass of block = 190 g = 0.190 kg, g = acceleration due to gravity = 9.8 m/s² and h = vertical height block moves and
K' = final kinetic energy = 0 (both kinetic energies are zero since the body is at rest at both positions)

Substituting the values of the variables into the equation, we have

U + P + K = U' + P' + K'

1/2kx² + 0 + 0 = 0 + mgh + 0

1/2kx² = mgh

Vertical height block moves

making h subject of the formula, we have

h = kx²/2mg

substituting the values of the variables into the equation, we have

h = kx²/2mg

h = 1.60 × 10³ N/m(0.1 m)²/(2 × 0.190 kg × 9.8 m/s²)

h = 1.60 × 10³ N/m × 0.01 m²/(3.724 kgm/s²)

h = 16.0 Nm/(3.724 kgm/s²)

h = 4.296 m

Distance up the incline the block moves

Now, the distance up the incline the block moves is L = h/sinФ where

h = vertical height block moves = 4.296 m and
Ф = angle of incline = 60.0°

Substituting the values of the variables into the equation, we have

L = h/sinФ

L = 4.296 m/sin60.0°

L = 4.296 m/0.8660

L = 4.96 m

So, how far up the incline (in m) the block moves from its initial position before it stops is 4.96 m

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