rules : everyone must attempt to answer each question

Rules:

1. Everyone must attempt to answer each question.

2. Show your work on a separate sheet of paper.

3. Keep track of your points.

4. Winners receive prize.

continue

Motion

Projectiles

Newton

ForcesVector

s

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2) Displacement = 27.4m + 3m -3m Distance = 27.4m + 3m +3m

continue

2) This object in freefall accelerates at 9.8 m/s/s. Every second the velocity changes by 9.8 m/s

continue

A ball rolls down an incline from rest. How far does it travel in 3 seconds if it accelerates at 5 m/s2?

Show Work for Creditd = vit + ½ at2

d = 0 + ½ (5 m/s2) (3s)2

d = 22.5 m continue

continue

4) Vavg = (Vf + Vi)/2 Vavg = (10m/s +25m/s)/2 Vavg = 17.5 m/s

1) d = vit + ½ at2

-6m = ½ a (2.7s)2

a = 1.6 m/s/s downcontinue

1) The acceleration of a projectileis always 9.81 m/s2 downward because the only force acting on the object is gravity. continue

2) d = vit + ½ at2

-22m = ½ a (3.0s)2

a = 4.9 m/s/s downcontinue

continueShow Work for Credit

VERTICALd = vit + ½ at2

-45m = 0 + ½ (-9.81m/s2) t2

t = 3.0 s

HORIZONTALd = vit + ½ at2

d = (25 m/s)(3.0 s) + 0d = 75 m

2) v vs. t graphs are

always linear.

Slope = accel. =

-9.81 m/s2

continue

vf = vi + at0 = 98 m/s + (-9.81m/s2)tt = 10. s

continue

continue

3) g = Fg/m

Fnet = ma12 N = (2kg+1kg)aa = 4 m/s2

continue

Which graph(s) depict an object that is accelerating?

B and C only

continue

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A B C D E

2) An object in equilibrium has an acceleration equal to zero. continue

Newton’s 3rd Law: the forces are equal in magnitude, opposite in direction.

Fnail on hammer = - Fhammer on nailcontinue

Fnet = ma50N – Ff = (4 kg)(10 m/s2)Ff = 10N

continue

Fnet = ma

Fapplied – Ffriction = 0

Fapplied = Ffriction

Fapplied = μFN = (0.05)(80kg x 9.81m/s2)

Fapplied = 40N

continue

continue

(C) If the net force is zero, the acceleration must also be zero.The slope of the speed graph = accel. = 0. Constant speed = constant slope of distance graph.

Fnet = ma

Fapplied – Ffriction = 0

Fapplied = Ffriction

Fapplied = μFN = (0.67)(60 N)

Fapplied = 40N

continue

Show Work for Credit

continue

Fnet = ma = 0

F – Ffriction – Fparallel = 0

F = Ffriction + Fparallel

F = Ffriction + Fg sinθ

F = 3N + (10N)sin30°

F = 3N + 5N

F = 8N

(4)Add vectors head to tail

continue

B

List 3 scalar quantities and 3 vector quantities. Include units.

Scalar: distance (m), speed (m/s), mass (kg), time (s),temp

(°C)

Vector: displacement (m), velocity (m/s), acceleration

(m/s2), Force (N), impulse (N·s), momentum (kg·m/s)continue

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to the question.

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have

continue

(4)FH = FcosθFH =(24N)cos35°FH = 20. N

FV = FsinθFV =(24N)sin35°FV = 14. N

continue

(1)

B

A R

continue

(4) Arrange 30N vectors to be head-to-tail, resultant starts at the tail of the first and ends at the head of the last.

R

30N

30N