8. 30 points OSColPhys1 7.P.001.Tutorial.WA. My Notes Ask Your Teacher As shown

Question

8. 30 points OSColPhys1 7.P.001.Tutorial.WA. My Notes Ask Your Teacher As shown in the figure below, a box of mass m-65.0 kg (initially at rest) is pushed a distance d-83.0 m across a rough warehouse floor by an applied force of FA-214 N 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.) 30° rough surface (a) work done by the applied force (b) work done by the force of gravity (c) work done by the normal force Wy= (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. WNet Tutorial

Explanation / Answer

normal force
Fn = mg + Fsin = 65.0kg * 9.8m/s² + 214N * sin30.0º
Fn = 744 N
so the friction force
f = µ*Fn = 0.100 * 744N = 74.4 N

(a) applied work = Fcos * d = 214N * cos30º * 83.0m = 15,382.34 J

(b) gravity is perpendicular to the direction of motion, so gravity work = 0 J

(c) same here; normal work = 0 J

(d) friction work = -74.4 N * 83.0m = -6175.2 J
(negative because it is opposed to the direction of motion)

(e) net work = 15,382.34 - 6175.2 = 9207.14 J

(f) net work = (Fcos - f) * d = (214N*cos30º - 74.4N)*83.0m = 9207.14 J

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