year 12 physics gradstart
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Year 12 Physics Gradstart. 2.1 Basic Vector Revision/ Progress Test. You have 20 minutes to work in a group to answer the questions on the Basic Vector Revision/Progress Test Worksheet. 2 .2 Vector Components Part 2. 2.2 Vector Components Part 2. - PowerPoint PPT PresentationTRANSCRIPT
Year 12 Physics Gradstart
2.1 Basic Vector Revision/Progress Test
You have 20 minutes to work in a group to answer the questions on the
Basic Vector Revision/Progress Test Worksheet
2.2 Vector Components Part 2
2.2 Vector Components Part 2
1a) What is the northerly and easterly component of the velocity below
vN = 35sin28 = 16.43150ms-1
E
vE = 35cos28 = 30.90317ms-1
2.2 Vector Components Part 2
1b) What is the northerly and easterly component of the velocity below
vN = 12cos15 = 11.59111ms-1
E
vE = – 12sin15 = – 3.10583ms-1
2.2 Vector Components Part 2
2. Work out the horizontal and vertical components of the force below:
y
x
Fx = 6cos30 = 5.19615N
Fy = 6sin30 = 3.0N
2.2 Vector Components Part 2
3. Work out the horizontal and vertical velocity components of the golfball below:
y
x
vx = 30cos60 = 15ms-1
vy = 30sin60 = 25.98076ms-1
2.2 Vector Components Part 2
4(a) Work out the weight into the slope and down the slope.
y
x
Wx = 15sin22 = 5.61910N
Wy = 15cos22 = 13.90776N W=mg
= 15N
22o
2.2 Vector Components Part 2
4(b) What is the acceleration down the slope if it is frictionless.
In x directiona = ? Fnet = 5.61910N m = 1.5kg
Fnet = a = a = a = 2.80955 a 2.8 ms-2
y
x
Wx = 15sin22 = 5.61910N
Wy = 15cos22 = 13.90776N W=mg
= 15N
22o
2.2 Vector Components Part 2
y
x
Wx = 15sin22 = 5.61910N
Wy = 15cos22 = 13.90776N W=mg
= 15N
4(c) What is the normal reaction force on the mass.
In y direction N = Wy
N = 13.90776NN 14N
N
22o
2.2 Vector Components Part 2
5(a) Work out an expression for the net force down the frictionless slope below.y
x
Wx = mgsin
W=mg
Fx = mgsin
2.2 Vector Components Part 2
5(b) What is the formula for the acceleration of a mass down a frictionless slope?
In x directiona = ? Fnet = mgsin m = m
Fnet = a = a = g sin
y
x
Wx = mgsin
W=mg
Fx = mgsin
2.2 Vector Components Part 26 If the surface below is frictionless, how long will it take the
mass to reach the base of the slope if it starts from rest?
t = ? u = 0 x = 0.4m a = g sin20o
= 10sin20o
= 3.40201x = ut + ½ at2
0.4 = ½ × 3.40201 × t2
0.233904 = t2
0.483637 = tt = 0.48m
A
Save in memory A of your calculator
2.2 Vector Components Part 27 If Frmax = 4.0N and a 15N is applied to the 3.0kg block at 30o to the
horizontal (a) What is the acceleration on the mass?
In x directionFnet = Fx - Frma = 12.99038 - 43a = 8.99038a = 2.99679a 3.0 ms-2
y
x
Fx = 15cos30 = 12.99038N
Fy = 15sin30 = 7.5NFrmax = 4.0N
B
It is not worth saving 7.5 in the
memory since it is so easy to enter in
the calculator
2.2 Vector Components Part 27 If Frmax = 4.0N and a 15N is applied to the 3.0kg block at 30o to the
horizontal (b) What is the normal reaction on the mass?
In y directionN + Fy = WN + 7.5 = 30N = 22.5NN 23N
Fx = 15cos30 = 12.99038N
Fy = 15sin30 = 7.5NFrmax = 4.0N
y
x
W = 3 × 10 = 30N
N
2.2 Vector Components Part 28 The 2.0kg mass below is held at rest on the 40o slope below by
friction. What is the magnitude of the friction holding the mass?
In x directionFr = Wx
Fr = 12.85575 NFr 13N
y
x
Wx = 20sin40 = 12.85575 NW=mg
= 20N
Fr
40o
2.2 Vector Components Part 29 What will be the acceleration of the 1.6kg block
on the 30o slope below if Frmax = 3.0N?
In x direction Fnet = Wx – Fr
ma = 8 – 31.6a = 5a = 3.125a 3.1 ms-2
y
x
Wx = 16sin30 = 8 NW=mg
= 16N
Frmax = 3.0N
40o
2.2 Vector Components Part 210 If Frmax = 2.0N and a 12N is applied to the 1.0kg block at 30o to the
horizontal (a) What is the acceleration on the mass?
In x directionFnet = 12 – Wx – Frma = 12 – 2 – 51a = 5a = 5.0 ms-2
yx
Wx = 10sin30 = 5 N
Wy = 10cos30 = 8.66025N
W=mg = 10N
Frmax = 2.0NA
30o
2.2 Vector Components Part 210 If Frmax = 2.0N and a 12N is applied to the 1.0kg block at 30o to the
horizontal (b) What is the normal reaction on the mass?
In y directionN = 8.66025NN 8.7N
Wx = 10sin30 = 5 N
Wy = 10cos30 = 8.66025N
W=mg = 10N
Frmax = 2.0NA
yxN
30o
2.2 Vector Components Part 211 If Frmax = 8.0N and a 20N is applied to the 3.0kg block at 30o to the
horizontal What is the acceleration on the mass?
In x directionThe force up the slope is not enough to overcome friction and the weight force down the slope (max = 23N)so the mass will be stationary.
yx
Wx = 30sin30 = 15 NW=mg
= 30N
Frmax = 8.0N
30o
2.2 Vector Components Part 212 A 55kg cyclist is riding up a 4.0o slope at a constant speed of 3.0ms-
1 ? (a) What is the net force on the cyclist?
Since the cyclist is travelling at constant speed the net force is zero.
2.2 Vector Components Part 212 A 55kg cyclist is riding up a 4.0o slope at a constant speed of
3.0ms-1 ? (b) What is the horizontal force of the rear wheel propelling the
cyclist up the hill if there is no rolling resistance (friction)?
In x directionsince Fnet = 0
F = 38.36606 NF 38N
Wx = 550sin4 = 38.36606 NW=mg
= 55 × 10 = 550N
F
4o
y
x
2.5 Classic Lift Problems
2.5 Classic Lift Problems1 If the lift below is moving vertically at a constant speed of 3.0 ms-1: (a) Draw a force diagram for the 50kg mass.
kg
W = 500N
T
2.5 Classic Lift Problems1 If the lift below is moving vertically at a constant speed of 3.0 ms-1: (b) What is the Normal Reaction on the mass?
Since the lift is travelling at constant speed the net force is zero and so
T = WT= 500Nkg
W = 500N
T
2.5 Classic Lift Problems1 If the lift below is moving vertically at a constant speed of 3.0 ms-1: (c) How much would the mass register on a set of scales?
Since apparent weight = TApparent Mass = Apparent Mass = Apparent Mass = 50kg
kg
W = 500N
T
2.5 Classic Lift Problems2 If the lift is accelerating upwards at 2.0ms-2: (a) Draw a force diagram for the 50kg mass.
kg
W = 500N
T
2.5 Classic Lift Problems2 If the lift is accelerating upwards at 2.0ms-2 : (b) What is the Normal Reaction on the mass?
Fnet = T – W ma = T - 50050 × 2 = T – 500 100 = T – 500 600 = T T = 600N
kg
W = 500N
Ta = 2.0ms-2 +
2.5 Classic Lift Problems2 If the lift is accelerating upwards at 2.0ms-2 : (c) How much would the mass register on a set of scales?
Since apparent weight = TApparent Mass = Apparent Mass = Apparent Mass= 60kg
kg
W = 500N
Ta = 2.0ms-2 +
2.5 Classic Lift Problems3 If the lift is moving upwards and decelerates at 3.0ms-2: (a) Draw a force diagram for the 50kg mass.
kg
W = 500N
T
2.5 Classic Lift Problems3 If the lift is moving upwards and decelerates at 3.0ms-2 : (b) What is the Normal Reaction on the mass?
Fnet = T – W ma = T - 50050 × -3 = T – 500 -150 = T – 500 450 = T T = 450N
kg
W = 500N
T
a = 3.0ms-2
+
2.5 Classic Lift Problems3 If the lift is moving upwards and decelerates at 3.0ms-2 : (c) How much would the mass register on a set of scales?
Since apparent weight = TApparent Mass = Apparent Mass = Apparent Mass= 45kg
kg
W = 500N
T+
a = 3.0ms-2