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A 20-kg block on a horizontal surface is attached to a light spring (force constant = 8.0 kN/m). The block is pulled 10 cm to the right from its equilibrium position and released from rest. When the block has moved 2.0 cm toward its equilibrium position, its kinetic energy is 13 J. How much work is done by the frictional force on the block as it moves the 2.0 cm

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The work done by the frictional force on the block as it moves the 2.0 cm is 6.44 J. It is negative because the friction opposes the motion of the block, dissipating energy in the form of heat.

How can you solve how much work is done by the frictional force on the block as it moves the 2.0 cm?

We have a mass (m = 20 kg) attached to a spring with a force constant (k = 8.0 kN/m).

The block is initially displaced from its equilibrium position by 10 cm (0.1 m) and then released.

We are interested in the work done by friction when the block moves 2 cm (0.02 m) towards its equilibrium position.

The total mechanical energy of the system remains constant unless acted upon by non-conservative forces like friction. Therefore:

Initial potential energy + work done by the spring = kinetic energy + work done by friction

E pi = 1/2 * k * x² = 1/2 * 8.0 kN/m * (0.1 m)² = 4.0 J

W spring = -ΔE pi = -(-4.0 J + E pi at 2 cm)

W friction = E pi + W spring - kinetic energy

W friction = 4.0 J + (-(-4.0 J + E pi at 2 cm)) - 13 J

W friction = E pi at 2 cm - 9 J

The block has moved 2 cm closer to the equilibrium position from its initial displacement of 10 cm. Therefore, its new displacement is 0.1 m - 0.02 m = 0.08 m.

E pi at 2 cm = 1/2 * k * (0.08 m)² = 2.56 J

W friction = 2.56 J - 9 J = -6.44 J

Therefore, the work done by the frictional force on the block as it moves the 2.0 cm is 6.44 J. It is negative because the friction opposes the motion of the block, dissipating energy in the form of heat.

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