Answer:
B. 560 J
J. 1.2 m
Explanation:
v = Final velocity = 0
u = Initial velocity = 4 m/s
[tex]\mu[/tex] = Coefficient of friction = 0.7
m = Mass of runner = 70 kg
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Kinetic energy is given by
[tex]K=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow K=\dfrac{1}{2}\times 70\times (0^2-4^2)\\\Rightarrow K=-560\ \text{J}[/tex]
The mechanical energy lost is 560 J
Acceleration is given by
[tex]a=-\mu g\\\Rightarrow a=-0.7\times 9.81\\\Rightarrow a=-6.867\ \text{m/s}^2[/tex]
From kinematic equations we get
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-4^2}{2\times -6.867}\\\Rightarrow s=1.165\approx 1.2\ \text{m}[/tex]
The runner slides for 1.2 m
a. The amount of mechanical energy that is lost because of friction acting on the runner is: B. 560 J
b. The distance the runner slide is equal to: J. 1.2 m
Given the following data:
Acceleration due to gravity, g = 9.8 [tex]m/s^2[/tex]
a. To determine the amount of mechanical energy that is lost because of friction acting on the runner, we would calculate the change in kinetic energy:
Mathematically, the change in kinetic energy is given by the formula:
[tex]\Delta K.E = \frac{1}{2} m(v-u)^2[/tex]
Substituting the parameters into the formula, we have;
[tex]\Delta K.E = \frac{1}{2} \times 70(4-0)^2\\\\\Delta K.E = 35(4)^2\\\\\Delta K.E = 35 \times 16\\\\\Delta K.E = 560\;Joules[/tex]
Mechanical energy = 560 Joules
b. To determine how far (distance) the runner slide:
First of all, we would find the runner's acceleration.
[tex]Acceleration = ug\\\\Acceleration = 0.70 \times 9.8[/tex]
Acceleration = 6.86 [tex]m/s^2[/tex]
For distance:
[tex]V^2 = U^2 + 2aS\\\\4^2 = 0^2 + 2(6.86)S\\\\16=13.72S\\\\S=\frac{16}{13.72}[/tex]
Distance, S = 1.2 meters.
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