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
Answer: 52.
Explanation: Critical Thinking When a ball is thrown vertically up, it keeps going up until it hits a certain position, and then falls down. At the maximum point, the velocity is instantaneously zero.
I may or may not be a nerd lol.
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Answer:
GRAVITATIONAL FORCE
Explanation:
We may have noticed that a body thrown upward in air falls back down again after attaining a particular height. The object was able to fall down back due to the effect of gravity acting on it. If there are no force of gravity acting on the body, the body will not fall back but rather disappears into the thin air.
A coin tossed upward in the air which falls back down when released is therefore under the influence of gravity i.e GRAVITATIONAL FORCE while it moves upward after it is released
Line c is at rest . line a is going in a positive direction . line b is going in a negative direction . line d is negative too
Answer:
K = -½U
Explanation:
From Newton's law of gravitation, the formula for gravitational potential energy is;
U = -GMm/R
Where,
G is gravitational constant
M and m are the two masses exerting the forces
R is the distance between the two objects
Now, in the question, we are given that kinetic energy is;
K = GMm/2R
Re-rranging, we have;
K = ½(GMm/R)
Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;
K = -½U