The frictional force is given by F = μmg
<span>where μ is the coeficient of friction. </span>
<span>Work done by frictional force = Fd = μmgd </span>
<span>Kinetic energy "lost" = 1/2 mv² </span>
<span>Fd = μmgd = 1/2 mv² </span>
<span>The m's cancel μgd = v² / 2 </span>
<span>d = v² / 2μg </span>
<span>d = 8² / 2(0.41)(9.8) </span>
<span>d = 32 / (0.41)(9.8) </span>
<span>d = 7.96 </span>
<span>Player slides 8 m . </span>
<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>
<span>d = v² / 2μg </span>
<span>= 4² / 2(0.46)(9.8) </span>
<span>= 8 / (0.46)(9.8) </span>
<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D
Mass and distance are the two factors
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
C, N and O all belong to the same period, in which it's 2nd Period.
