A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distance d. When the block is released, it slides across a surface that is frictionless except for a rough patch of width 5.0 cm that has a coefficient of kinetic friction = 0.44. Find d, such that the block's speed after crossing the rough patch is 2.3 m/s.

Compression distance:

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy (), in joules, at the expense of elastic potential energy (), in joules, and overcoming work losses due to friction (), in joules:

(1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

(2)

Where:

- Mass of the block, in kilograms.

- Final velocity of the block, in meters per second.

- KInetic coefficient of friction, no unit.

- Gravitational acceleration, in meters per square second.

- Width of the rough patch, in meters.

- Spring constant, in newtons per meter.

- Compression distance, in meters.

If we know that , , , , and , then the compression distance of the spring is:

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation:

d is the correct answer