My Notes . Ask As shown in the figure below, a box of mass m=62.0 kg (initially

Question

My Notes . Ask As shown in the figure below, a box of mass m=62.0 kg (initially at rest) is pushed a distance d=66.0 m across a rough warehouse floor by an applied force of FA-234N directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.) ouh surface (a) work done by the applied force (b) work done by the force of gravity Wg- (c) work done by the normal force (d) work done by the force of friction (e) Calculate the net work on the box by finding the sum of all the works done by each individual force. (f) Now find the net work by first finding the net force on the box, then finding the work done by this net force. Additional Materials Reading

Explanation / Answer

Given,

m = 62 kg ; d = 66 m ; Fa = 234 N ; theta = 30 deg ; uk = 0.1

a)We know that

W = F d cos(theta)

W = 234 x 66 x cos30 = 1.34 x 10^4 J

Hence, W = 1.34 x 10^4 J

b)Wg = 0 J

c)Wn = 0 J

d)Wf = uk N d

N = mg + Fa sin(theta) = 62 x 9.81 + 234 x sin30 = 725.22 N

Wf = 0.1 x 725.22 x 66 = 4.79 x 10^3 J

Hence, Wf = - 4.79 x 10^3 J

e)Wnet = 1.34 x 10^4 - 4.79 x 10^3 = 8.61 x 10^3 J

Hence, Wnet = 8.6 x 10^3 J

f)Fnet = Fapp - Fric = Fa cos30 - uk N

Fnet = 234 x cos30 - 0.1 x 725.22 = 130.13 N

Wnet = Fnet d = 130.13 x 66 = 8.6 x 10^3 J

Hence, Wnet = 8.6 x 10^3 N

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