Answer:
a) W₁ = 8242.2 J, b) W₁ = 8242.2 J
, c) W₃ = 0
, d) W₄ = -189.51 J ,
f) v = 27.24 m / s
Explanation:
a) Work is defined by
W = F. d = F d sin θ
where angle is between force and displacement
n this case the suitcase is going up and the outside F is parallel to the plane, so the angle is zero and the cosine is 1
W = F d
Let's calculate
W = 169 3.8
W₁ = 8242.2 J
b) the gravitational force is vertical so it has an angle with respect to the horizontal parallel to the plane of
θ’= 90 - θ
θ'= 90-32 = 58º
W = m g d thing θ ’
W = 20 9.8 3.8 thing (180 + tea ’) =
W = 744.8 cos (180 + 32)
W₂ = -631.6 J
c) The normal work, as it has 90º with respect to the displacement, its work is zero
W₃ = 0
d) the work of the friction force
Let's write Newton's second law the Y axis
N- Wy = 0
Cos 32 = Wy / W
N = W cos 32
The expression for friction force is
fr = μ N
fr = μ mg cos 32
fr = 0.300 20 9.8 cos (32)
fr = 49.87 N
The work of the friction force
W = fr d cos 180
W₄ = -49.87 3.8
W₄ = -189.51 J
E) The total work
W = W₁ + W₂ + W₃ + W₄
W = 8242.2- 631.6 + 0 -189.51
W_total = 7421.09 J
F) Usmeosel theorem of work and energy
W = ΔK
W = ΔK = ½ m v² - 0
v =√ 2W / m
v = √ (2 7421.09 / 20)
v = 27.24 m / s
