Answer:
(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
