A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/s by pulling on the rope tied to the crate with a force of 50 N. The rope makes an angle of 25° with the horizontal.

a. What are the horizontal and vertical components of the applied force?

b. What is the magnitude of each of the forces? i) Applied Force ii) Weight ii) Normal Force iii) Frictional Force

c. How much work is done by each of the forces? i) WApplied Force ii) WWeight iii) WNormal Force iv) WFrictional Force

d. What is the total amount of work done on the object?

e. What is the coefficient of friction of the crate on the floor?

A) Fx = 45.3N Fy = 21.1N

B) Fm = 45.3N R = 196N Ff = 35.4N

C) i) 0KJ ii) 11.5KJ iii) 11.5KJ iv) 0KJ

D) 23KJ

E) 0.047

Explanation:

a) Horizontal component of the force Fx = 50cos 25° = 45.3N

Vertical component = 50sin25° = 21.1N

b) Magnitude of the applied force = moving force Fm = Fx = 50cos25° =45.3N

Magnitude of the weight = mg = 20×9.8 = 196N

Magnitude of normal force which is the reaction is equal to the weight = mg = 20×9.8 = 196N

Frictional force = moving force = 50cos45° = 35.4N

c) since workdone = Force done × perpendicular distance in the direction of the force

- workdone on the moving force is 0Joules since it has no perpendicular distance

- workdone on weight is the weight × distance = (20×9.8)×12 = 11.5KJ

= work done on normal force = workdone by the weight = 11.5KJ

Workdone on the frictional force is is 0Joules since the force is along the horizontal (no perpendicular distance)

d) the total work done = work done by Applied Force +Weight + Normal Force + Frictional Force= 0+11.5+11.5 = 23KJ

e) the coefficient of friction = moving force / normal reaction = 45.3/196 = 0.047