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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= (vfvi)/(tfti) 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 = 8J3J= 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 