Part II 0.400 and a) Consider an object of Mass M = 7.50 kg object placed on a h

Question

Part II 0.400 and a) Consider an object of Mass M = 7.50 kg object placed on a horizontal rough surface ( 0.300) initially at rest. A force F is applied to the object, where the force is directed upward 20.0 above the horizontal. The magnitude of the force is 28.5 N. Find the speed and displacement of the object 4.55 s after the force F is applied Simulate on IP and determine the % error in the speed and the displacement. b) Repeat the above problem, but the applied force F is now directed 20.0 below the horizontal. Find the speed and displacement of the object 2.00 s after the force F is applied.

Explanation / Answer

a)Mass of object=M=7.5 kg

applied force=28.5N

angle with horizontal=20o

Horizontal component of force applied = 28.5*cos20=26.78N

normal force from surface on object=(7.5*9.8)-(28.5sin20)=63.75N

Static Maximum static friction that can develop=0.4*63.75=25.5N

Since the horizontal component of applied force is higher that static friction force,the block will move.

Friction force on block=0.3*63.75=19.125N

Net force in horizontal direction on object=26.78-19.125=7.655N

acceleration of object=7.655/7.5=1.02 m/s2

speed after 4.55 seconds = 1.02*4.55=4.64m/s

Displacement =(1/2)*1.02*4.552=10.56m

b)If the force direction is 20o below horizontal, normal force on block from surface=(7.5*9.8)+(28.5sin20)=83.25N

Maximum static friction that can develop=0.4*83.25=33.3N

Applied force in horizontal direction is less than the friction force.Therefore, the object will not move at all in this case.

Speed=displacement =0 after 2 seconds

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