
Newton’s three laws describe how motion is affected by forces. The first law says that, in the absence of a net force, an object at rest or in motion will continue to be in the identical state of rest or motion (that is, it will continue to have the same velocity). The second law states the relationship among net force, mass, and acceleration experienced by an object. The third law states that forces occur in pairs: For every “action” force (on a particular object), there is an equal but opposite “reaction” force operating at the same time, but on another object.

friction, Newton’s first law of motion, Newton’s second law of motion, Newton’s third law of motion, reaction force


 Review problems and questions 

 Which of Newton’s three laws of motion best applies to each statement?
 A 120 lb student sat on a stool, and the stool applied a 120 lb normal force to the student.
 The spacecraft cruised toward the North Star at a constant 500 m/s through the vacuum of interstellar space.
 Now that the moving van was full, its driver had to apply more force to reach highway speed before the onramp ended.

 Newton’s third law. The force of the student pressing down on the stool and the normal force from the stool pressing up on the student are an action–reaction pair of forces.
 Newton’s first law. In the absence of a net force, the spacecraft remains in a constant state of motion.
 Newton’s second law. With more mass, the van needs more force to achieve the same acceleration.

 Two students are debating the meaning of Newton’s laws of motion.
Arjun: “See that guy pushing the crate across the gym floor? He’s working hard, but the crate never speeds up. No acceleration! Newton’s second law can’t be right!”
Buell: “I wonder. Newton’s second law has to do with net force, which is the total force you are left with after you add up all the forces pushing on something. I think there is another force at work.”
Which student is correct, and why?

Buell is correct. Newton’s second law states that an object’s acceleration is proportional to the net force acting on an object. In the direction of the crate’s motion, the worker is pushing just hard enough to overcome friction between the crate and the floor—but no harder. (He did have to push harder in order to start the crate moving.) In the vertical direction, the crate’s weight equals the normal force applied by the floor. Sure enough, the crate is not accelerating along the vertical axis either! (It remains at the level of the floor.)

 Rank the following objects by the magnitude of the net force that each experiences, from weakest to strongest:
 150 g hockey puck that leaves a player’s stick at 30 m/s and enters the goal one second later at 29 m/s
 20 g marble that has been dropped 1 m above the floor (and is still falling)
 golf ball that is tapped with a 0.5 N force while resting on a table

From weakest to strongest: a, b, c First, consider the hockey puck.
Asked: net force on hockey puck Given: puck mass m = 0.15 kg; initial velocity v_{0} = 30 m/s; final velocity v_{f} = 29 m/s; time interval Δt = 1 s Relationships: F_{net} = ma; a = (v_{f} − v_{0})/Δt Solution: We know the puck’s mass, so we need to calculate its acceleration. $$a=\frac{({v}_{f}{v}_{0})}{\Delta t}=\frac{(2930)\text{m/s}}{\text{1s}}=\text{1}\text{m}/{\text{s}}^{2}$$Now that we know the puck’s acceleration, we can calculate the net force upon it. $${F}_{net}=ma=(0.15\text{kg})(1\text{}\text{m}{\text{s}}^{\text{2}})=0.15\text{N}$$The puck experiences a net force of −0.15 N; the magnitude (or strength) of that force is 0.15 N. (Note that this is roughly onetenth of the puck’s weight, which is given by F_{w} = mg = 1.47 N. This indicates relatively little friction between the puck and the ice.)
 Next, consider the marble, which feels only the force of its weight because it is in free fall.
Asked: net force on the marble Given: marble mass m = 0.02 kg; acceleration a = g = 9.8 m/s^{2} Relationship: F_{net} = ma Solution: $${F}_{net}=ma=(0.02\text{kg})(9.8\text{}\text{m}/{\text{s}}^{\text{2}})=0.2\text{N}$$The marble experiences a net force of 0.2 N. Note that this is simply its weight! In fact, the formula for weight (F_{w} = mg) is a special case of Newton’s second law for an object in free fall.
 Finally, the net force on the golf ball is stated as 0.5 N.
 Take a Quiz 
 
