Newton's Laws of Motion Review

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Part E: Force-Mass-Acceleration Relationships


Construct free-body diagrams for the following objects; label the forces according to type. Use the approximation that g=10 m/s2
to determine the magnitude of all forces and the net force and acceleration of the object.

38. A 2-kg box is at rest on a table.

 

("At rest" would indicate a balance of forces and an acceleration of 0 m/s/s.)

39. A 2-kg box is free-falling from the table to the ground.

("Free-falling" indicates that the only force that influences the motion is the force of gravity.)

40. A 2-kg box equipped with a parachute is falling at a terminal velocity after being dropped from a plane.

(A "terminal velocity" indicates a constant velocity and a balance of forces.)



41. A 2-kg box is sliding to the right across a table. The force of friction upon the box is 5 N.

 

(Friction is directed opposite the motion and causes a leftward acceleration; no rightward force is spoken of, only a rightward motion.)

42. An 8-N force is applied to a 2-kg box to move it to the right across the table at a constant velocity of 1.5 m/s.

 

(A "constant velocity" indicates an acceleration of 0 m/s/s and a balance of forces.)


43. An 8-N force is applied to a 2-kg box to accelerate it to the right across a table. The box encounters a force of friction of 5 N.

(The horizontal forces can be summed as vectors; divide by the mass to obtain the acceleration value.) 


44. A 500-kg freight elevator is descending down through the shaft at a constant velocity of 0.50 m/s.

(A "constant velocity" indicates an acceleration of 0 m/s/s and a balance of forces.)  


45. A 500-kg freight elevator is moving upwards towards its destination. Near the end of the ascent, the upward moving elevator encounters a downward acceleration of 2.0 m/s/s.

 

(Begin by multiplying m•a to find the net force - 1000 N, down. The downward gravity force must be 1000 N more than the upward tension force.)


46. A 150-N rightward force is applied to a 20-kg box to accelerate it to the right across a rough surface at a rate of 2.0 m/s/s.

 

(Begin by multiplying m•a to determine the net force - 40 N, right. The rightward applied force must be 40 N more than the leftward friction force.)

 
Useful Web Links

Free-Body Diagrams || Finding Acceleration || Finding Individual Forces
 

#37 | #38 | #39 | #40 | #41 | #42 | #43 | #44 | #45 | #46 ]

 



 

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Answers to Questions:  All  ||  #1-7  ||  #8-36  ||  #37-46  ||  #47-60



 

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