The coefficient of kinetic friction (μ) between the block and the table is 0.4.
<h3>
What is kinetic friction?</h3>
This sis the frictional force between an object in motion with the surface in contact.
μN = ff
where;
- N is normal reaction due to weight of the block
- ff is frictional force
- μ is coefficient of friction
μ = ff/N
μ = 8/20
μ = 0.4
Thus, the coefficient of kinetic friction (μ) between the block and the table is 0.4.
Learn more about coefficient of friction here: brainly.com/question/20241845
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Answer:
The coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.
Explanation:
Since the block reads 24.5 N before the block starts to move, this is its weight. Now, when the block starts to move at a constant velocity, it experiences a frictional force which is equal to the force with which the student pulls.
Now, since the velocity is constant so, there is no acceleration and thus, the net force is zero.
Let F = force applied and f = frictional force = μN = μW where μ = coefficient of friction and N = normal force. The normal force also equals the weight of the object W.
Now, since F - f = ma and a = 0 where a = acceleration and m = mass of block,
F - f = m(0) = 0
F - f = 0
F = f
Since the force applied equals the frictional force, we have that
F = μW and F = 23.7 N and W = 24.5 N
So, 23.7 N = μ(24.5 N)
μ = 23.7 N/24.5 N
μ = 0.97
Since μ = 0.97 < 1, the coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.
<span>This question is based on conservation of energy as the work done would lead to change in kinetice energy of car
change in KE = 1/2 mv(f)^2 - 1/2mv(i)^2 = 1/2m(v(f)^2-v(i)^2)
where v(f) and v(i) are the final and initial speeds
change in KE = 185kJ = 185,000J = 1/2 m((28m/s)^2-(23m/s)^2)
185,000=1/2 m(255m^2/s^2)
solving for m
m=1451kg</span>
It looks blue as it is only reflecting blue light
It would be d all of the above
