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The concepts of mass, weight, and acceleration as they relate to falling objects and the force of gravity. Topics include the relationship between mass and inertia, the effect of gravity on objects, and the experimental evidence for the universality of free fall.
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Do Falling Objects Accelerate?• It sure seems like it!– Starts from rest, goes faster and faster....• What about a feather, though?– Air resistance, drag– Terminal velocity– What if we could get rid of the air?• What’s responsible for the downwards force?– If it’s accelerating, then a force is acting:^ F^ = m a
Golf Ball vs. Bowling Ball• Which one is more massive?• Which one experiences more gravitational force?• Which one is most reluctant to accelerate?• How do they respond to a gravitational force?
-^ Gravitational force is proportional to mass•^ F^ = m a^ gives an object’s responding acceleration•^ Divide both sides of the equation by “
-^ a^ =^ F /m •^ Both numerator and denominator are proportionalto “ m ”, if force is gravity•^ SO....acceleration is the same, regardless of themass•^ We’ll return to this point when we considerGeneral Relativity!
-^ Ignoring air resistance, falling objects near thesurface of the Earth experience a constantacceleration of 9.8 m/s
-^ That means if you drop something it goes fasterand faster, increasing its speed downwards by9.8 m/s in each passing second.
-^ A ball falls from rest for 4 seconds. Neglecting air resistance,during which of the 4 seconds does the ball’s speed increasethe most?•^ If you drop a ball from a height of 4.9 m, it will hit the ground1 s later. If you fire a bullet exactly horizontally from a heightof 4.9 m, it will also hit the ground exactly 1 s later. Explain.•^ If a golf ball and a bowling ball (when dropped from the sameheight) will hit your foot at the same speed, why does one hurtmore than the other?
TimeAcceleration^2 Interval(m/s^ down)
InitFinalAverageVelocityVelocity(m/s down)(m/s down) 0 – 1 s^10
1 – 2 s^10
2 – 3 s^10
3 – 4 s^10
4 – 5 s^10
Starting from rest, letting go:
The average velocity inthe interval is justV= ½(vavg^
TimeAcceleration^2 Interval(m/s^ down)
FinalAverageVelocityVelocity(m/s down)(m/s down)
Dist.FinalPositionmoved(m down)(m down)
0 – 1 s^10
1 – 2 s^10
2 – 3 s^10
3 – 4 s^10
4 – 5 s^10
Starting from rest, letting go: