chapter 5 review
DESCRIPTION
Chapter 5 Review. Work is force x distance. Only the force parallel to the displacement applies to the work. F pull = 20N. Pulling a box. N = 137.2N. f = -16.5N. F pully = sin30° 20N = 10N. F pull = 20N. 30°. angle = 30°. F pullx = cos30°(20N) = 17.3N. mass = 15 kg. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 5 Review
Work is force x distance
Only the force parallel to thedisplacement applies to the work
Pulling a box
Fpull = 20N
angle = 30°30°
mass = 15 kgpulled 8 m = .12
Label and calculate all forces
Fpull = 20N
Fpullx = cos30°(20N) = 17.3N
Fpully = sin30°20N = 10N
mg = -147.2N
N = 137.2Nf = -16.5N
Calculate the work done by the pulland friction
Wpull = Fxd = 138.4J
Wf = fd = -132J
Potential Energy is energy of position
GPE = mgh affected by height
EPE = ½kx2
affected by distance squeezed or pulled
k is the spring constant
Kinetic Energy is energy of motion
KE = ½mv2
affected by velocity
if I triple the speed of somethingwhat happens to it’s KE?
Mechanical Energy is PE + KE
MEi = MEf if there is no friction
if there is friction, some KEis changed to heat and lost
40 J of work is done toraise the ball of a pendulum
parallel to the ground30°
If it is released how much KE,PE and ME will it have after 30°?
At its lowest point, how much KE,PE and ME does it have?
½h
½h
10 m
3 m/s
What will the wagon’s speedbe at the bottom of the hill?
KEi + PEi = KEf
½mv2i + mghi = ½mv2
f
v2f = v2
i + 2ghi
vf = 14.3 m/s
5 kg
A 5 kg block is pushed 20 cmagainst a spring with a springconstant of 240 N/m
What is the stored elastic potential energy?
EPE = ½kx2 = ½(240N/m)(.2 m)2 = 4.8 J
If released, what is the velocity of the block, as it comesoff the spring?
KE = EPE = 4.8 J = ½mv2 v = 1.39 m/sIf 4 N of frictional force is acting between the floor andblock, what distance will the block go?
W = KE = f d d = KE / f = 4.8 J / 4 N= 1.2 m