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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


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?


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?


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?


Can only multiply and divide, so act as a coefficient

-1X 3N= -3N

2x2N= 4 N

Work definition



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


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)=


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?


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




The rate that work is done

1J/per second= 1 Watt



Power = (mgh)/time

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

Power= mgh/t


need seconds

55x9.8x1300/ 7200=

97.3 Watts or Joules/sec




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