Answer:
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
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
[tex]\frac{1}{2} kx^2 = Fd + \frac{1}{2} mv^2[/tex]
First, we need to get the initial velocity of the body using the equation of motion as shown:
[tex]\triangle v_x = v_0t + \frac{1}{2} at^2[/tex]
Note that the velocity in the y-direction is expressed as:
[tex]\triangle v_y = v_0t + \frac{1}{2} at^2[/tex]
Δy = 1.00 m
v₀ = 0 m/s
a = 9.8 m/s²
[tex]1 = 0t + \frac{1}{2}( 9.80t^2)\\1=4.9t^2\\t^2=\frac{1}{4.9} \\t^2=0.2041\\t=0.452secs[/tex]
Similarly in the x-direction, recall that
Δx = 0.20 m
a = 0 m/s²
t = 0.452 s
Get the initial velocity v₀ ;
[tex]0.2 = v_0(0.452) + \frac{1}{2} 0(0.452)^2\\0.2=0.452v_0+0v_0=\frac{0.2}{0.452}\\v_0= 0.442m/s[/tex]
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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