Answer:
I do belive that it is B hrs cn I an gn
By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)
D. Budgeting time, avoiding stress, and prioritizing.
Ek = 1/2 mv^2
9 × 10^4 = 1/2 × 800 × v^2
9 × 10^4/400 = 400 v^2 / 400
9 × 10^4/400 = v^2
√225 = v
15 ms⁻¹ = v
That's the only way I know how to work it out
I think in this case velocity and speed would be considered the same because me
s = d/t and v=d/t
one is distance travelled and the other is displacement of a body
Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.
