An m = 20.0 g object is held against the free end of a spring of constant k = 25.0 N / m that is compressed a distance x = 10.0 cm from its equilibrium length. Once released, the object slides d 1 = 1.15 m across the tabletop and eventually lands d 2 = 0.2 m from the edge of the table on the floor, as shown in the figure. Calculate the coefficient of friction μ between the table and the object. The sliding distance includes the compression of the spring, and the tabletop is h = 1.00 m above the floor level.

The coefficient of friction μ between the table and the object is 0.546

For us to get the coefficient of friction μ, we will use the law conservation of energy as shown.

Mathematically, Elastic energy = work + kinetic energy

First, we need to get the initial velocity of the body using the equation of motion as shown:

 Note that the velocity in the y-direction is expressed as:

Δy = 1.00 m

v₀ = 0 m/s

a = 9.8 m/s²

Similarly in the x-direction, recall that

Δx = 0.20 m

a = 0 m/s²

t = 0.452 s

Get the initial velocity v₀ ;

Next, use conservation of energy to find μ.

Elastic energy = work + kinetic energy

½ kx² = Fd + ½ mv²

½ (25.0) (0.100)² = (0.020× 9.8 × μ) (1.15 m) + ½ (0.020) (0.443 )²

On solving the expression for coefficient of friction, μ = 0.546

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μ = 0.546

Explanation:

First, use kinematics to find the object's velocity at the edge of the table.

Given in the y direction:

Δy = 1.00 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

1.00 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.452 s

Given in the x direction:

Δx = 0.20 m

a = 0 m/s²

t = 0.452 s

Find: v₀

Δx = v₀ t + ½ at²

0.20 m = v₀ (0.452 s) + ½ (0 m/s²) (0.452 s)²

v₀ = 0.443 m/s

Next, use conservation of energy to find μ.

Elastic energy = work + kinetic energy

½ kx² = Fd + ½ mv²

½ (25.0 N/m) (0.100 m)² = (0.020 kg × 9.8 m/s² × μ) (1.15 m) + ½ (0.020 kg) (0.443 m/s)²

μ = 0.546

answer: b.

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