A luggage handler pulls a 19.0-kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 169 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.330. The suitcase travels 3.90 m along the ramp.

a)Calculate the work done on the suitcase by F⃗ .

Express your answer with the appropriate units.

b)Calculate the work done on the suitcase by the gravitational force.

c)Calculate the work done on the suitcase by the normal force.

d)Calculate the work done on the suitcase by the friction force.

e)Calculate the total work done on the suitcase.

f)If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled3.90 m along the ramp?

(a) 659.1 J

(b_) 368.56 J

(c) 0 J

(d) - 202.72 J

(e) 824.94 J

(f) 9.32 m/s

Explanation:

mass, m = 19 kg

inclination of the plane, θ = 34°

Force, F = 169 N

coefficient of friction, μk = 0.330

distance moved, d = 3.90 m

(a)

Work done by the force on the suitcase

W = F x d = 169 x 3.90 = 659.1 Joule

(b)

Work done by the gravitational force on the suitcase

W = F x d x Sinθ

W = 169 x 3.9 x Sin 34°

W = 368.56 Joule

(c)

Work done by the normal force on the suitcase

W = Normal force x distance x Cos 90

As the angle between the normal force and the distance is 90°

W = 0 Joule

(d)

Work done by the friction force

W = friction force x distance x Cos 180°

W = - μk x mg x cos 34 x 3.9

W = - 0.33 x 19 x 10 x 3.9 x 0.829

W = - 202.72 Joule

(e)

Total work done

W = 659.1 + 368.56 + 0 - 202.72

W = 824.94 Joule

(f)

According to the work energy theorem, the net work done is equal to the change in kinetic energy of the body.

824.94 = 0.5 x 19 ( v² - 0)

v = 9.32 m/s

f is prop to m1xm2 double m1 double force

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray