Answer:
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
Explanation:
A.
Horizontal component, Fy = F * cos(a)
= 50cos25
= 50 * 0.91
= 45.32 N
Vertical component,Fx = F * sin(a)
= 50sin25
= 50 * 0.42
= 21.13 N
B.
Applied force = 50 N
Weight,W = m * g
= 20 * 9.81
= 196.2 N
Normal force, N = W - Fx
= 200 - 21.13
= 179 N
Frictional force = Fy
= 45.32 N.
C.
Workdone by Normal and Weight forces are = 0, because they both act perpendicular to the movement.
Workdone by friction = Workdone by applied forces
= force * distance
= 12 * 45.32
= 543.84 J
D.
Total amount of work done on the crate = 0 (the movement with a constant speed).
E.
The coefficient of friction, u = Fy/(W - Fx)
= 45.5/(196.2 - 21.13)
= 0.26
