Answer:
y = 12.82 m
Explanation:
We can solve this exercise using the energy work theorem
W = ΔEm
friction force work is
W = fr . s = fr s cos θ
the friction force opposes the movement, therefore the angle is 180º
W = - fr s
we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular
N -Wy = 0
N = mg cos θ
the friction force remains
fr = μ N
fr = μ mg cos θ
work gives
W = - μ mg s cos θ
initial energy
Em₀ = ½ m v²
the final energy is zero, because it stops
we substitute
- μ m g s cos θ = 0 - ½ m v²
s = ½ v² / (μ g cos θ)
let's calculate
s = ½ 20² / (0.55 9.8 cos 20)
s = 39.49 m
this is the distance it travels along the plane, to find the vertical distance let's use trigonometry
sin 20 = y / s
y = s sin 20
y = 37.49 sin 20
y = 12.82 m
Stan’s displaceent is 0 km because he ends his walk where he started.
Answer:
The distance of the student from the center of the merry-go-round is, r = 1.74 m
Explanation:
Given,
The acceleration of the student in merry go round, a = 3.6 m/s²
The tangential velocity of the student is, v = 2.5 m/s
The acceleration of the merry go round is given by the formula,
a = v² / r
Therefore,
r = v² / a
= 2.5² / 3.6
= 1.74 m
Hence, the distance of the student from the center of the merry-go-round is, r = 1.74 m
