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30 Cards in this Set

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Velocity= displacement/time




Average Velocity: (displacement for trip)/time

Average Velocity= 45m/3s= 15m/s or




1/2(vi+vf)= 1/2(30m/s+0)= 15m/s

Acceleration

rate of change of velocity or direction




A= (vf-vi)/(tf-ti)




change in velocity/change in time

If you are driving on a circular track at a constant speed, is acceleration zero?

No, because constantly changing direction, so that change is an acceleration

Xi = inches


vi = velocity in/s


a = accecleration in/s2

Newtons first law

object at rest stays are rest


in motion stays in motion

Newton's Second Law




units for force?




Weight?

Force is proportional to acceleration




Force = Mass x Acceleration




1 Newton= force to move 1kg by 1m/s2




Weight is in Newtons= kgx 9.8m/s2

Tennis Ball and bowling ball are dropped.




Does one have a greater force, neglecting air resistance?

Although the acceleration is the same, the bowling ball requires more force (Newtons) to accelerate down because




F= Mass kg x Acceleration in m/s2

What is constant acceleration?




If a ball is thrown straight up, is it still accelerating when it stops?

meaning constant force applied




F= Mass kg x Acceleration in m/s2




Yes, still accelerating

Can object move up or down with decreasing velocity?




Can speed be negative?

Yes




No, speed is positive, so if showing a speed curve on a ball thrown up, it slows to zero and speeds back up.

Is a ball resting on a table accelerating?

Yes, table is providing an opposite force.

Newtons Third Law





For every action, there is an equal and opposite reaction.

Example of a scalar?




How do we relate these to vectors?

Temperature




Can only multiply and divide, so act as a coefficient




-1X 3N= -3N


2x2N= 4 N

Work definition




Formula




units?

Force applied over distance




W= Force(N) x Displacement (M)=


so, Newtonxmeters, or


W= MxAxD= (kgxm)/s2 x (m)




Joules =Kgxm2/s2




only force in direction of displacement counts towards work

Kinetic Energy definition




Formula?




If you roll a 1kg ball on the floor and exert 6N over 0.2M, What is the kinetic energy?

Energy associated with the motion of an object


KE= Fxd for initial calculation= NxM=J


KE= (1/2)(mass)(v2)




Work is the kinetic energy, only use other formula to solve for final speed.


so KE= W= (kg already factored)6Nx0.2M= 1.2 J

0.1 kg ball


6N force


over 2M




What is the KE?




What is the final velocity?

KE = W= MAd= Fd= 6Nx0.2M= 1.2J




KE= 1/2mv2, so




1.2J= (1/2)(0.1kg)(v2)=




(1.2J)/(1/2)(0.1kg)= v2




=4.9 m/s

How do you calculate final velocity?

Plug total Joules in as KE for KE=1/2 mv2






and solve for V

4kg ball rolling on flat surface at 2 m/s




6N force is applied over 0.5m in opposite direction




initial kinetic energy?


Work by force?


Final Kinetic energy?


Final Velocity?

Initial KE= (1/2)4kgx (2m/s)2= 8J




work by forced= 6Nx0.5M= 3J




Final KE = 8J-3J= 5J




Vf? 5J= (1/2) 4kg (v2)=


(5J)/(1/2)4kg= v2= 1.6 m/s

Potential Energy

Same as calculating Work =FxD


where distance is height.


PE= mgh or


mass x acceleration x height





Calculate the potential energy of




2kg box on shelf 1.5m high

PE= mgh




PE= (2kg)(9.8m/s2)(1.5m)=




29J

Relate the Law of conservation of energy to Potential Energy

Energy cannot be created or destroyed, therefore 29J of Potential Energy can be converted to 29J of Kinetic Energy

What if 10J of energy is used to push a box along a surface? Does it have 10J PE?

No, 10J was lost in the form of heat energy or friction

4kg box 1.5m off the ground. What is its speed when it hits the ground?

PE= mgh


so - (4kg)(9.8m/s2)(1.5m)= 59J




Then KE=1/2mv2




59J= (1/2) 4kg v2


=5.4 m/s

Skier on top of 30 degree hill, what is speed at bottom?




55kg


100m high

PE= mgh


55kgx100mx9.8= 53900J= PE which is converted to KE


KE= 1/2mv2


53900J=1/2 (55kg) v2


Speed= 44.3 m/s




Easier way for velocity with gravity=


V= root(2gh) or V=root(2x9.8x100m)= 44.3 m/s

What if hill is bumpy or goes up or down

same speed because energy is conserved

55kg skier


100m hill




How much energy is lost to "other forces" if final velocity is 10m/s?

PE= mgh= 55x9.8x100m= 53,900J




KE= 1/2mv2 = 1/2 (55kg) (10m/s2)=2750J




So for loss, take PE at top- KE at Bottom


=51,150J lost to other forces



55kg skier already traveling at 10m/s


at top of 100m hill




What is their speed at bottom assuming no other factors are at play?

Total energy at top is PE + KE




So PE = mgh= 55x9.8x100= 53900


KE= 1/2 55kg x (10)2 = 2750




Total energy at top= 56,650


For speed KE =mv2


or 56650= 1/2(55kg)x (v)2= 45.4 m/s not much different

Power




Unit




Formula?

The rate that work is done




1J/per second= 1 Watt


Joules/second




Power=work/time




Power = (mgh)/time

average power output hiker 55kg who climbs 1300m in 2h

Power= mgh/t


55x9.8x1300/(2h)


need seconds




55x9.8x1300/ 7200=




97.3 Watts or Joules/sec



Momentum




sign?




definition

Momentum is vector quantitiy




signified by little p or p


Force = rate of change of momentum

momentum p




Momentum greater, 100kg man at 2m/s


or 2000kg car at 1 m/s?

p= Mass x Velocity




Man p = 100kg x 2 m/s= 200




Car p = 2000kg x 1 m/s = 2000




car is greater