A luggage handler pulls a 20.0 kgkg suitcase up a ramp inclined at 32.0 ∘∘ above the horizontal by a force F⃗ F→ of magnitude 169 NN that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is μkμk = 0.300. The suitcase travels 3.80 mm along the ramp. calculate:
A) the work done on the suitcase by force F.
B) the work done on the suitcase by the gravitational force.
C) the work done on the suitcase by the normal force.
D) the work done on the suitcase by the friction force.
E) the total work done on the suitcase.
F) if the velocity of the suitcase is zero at the bottom of the ramp, what is its velocity after it has travelled 4.60 m along the ramp.

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

power out to power in less than 100%

the entire statement is true and correct.  

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