Download Question 2 - RobboPhysics

UNIT 4.
PART F: Unit2 Revision
Question1
Vel-Time graph. Disp + Area under graph = ( 5 x 6 ) / 2 + 1 x ( 4 + 5 ) /2 = 19.5 m
Question 2
Forward: Frictional force of the road on the back tyres.
Backward: Air resistance on car, and frictional force of the road on the front tyres.
The forward vector should be greater than the sum of the backward vectors.
Question 3
200
Net Force
Question 4
Vertically: mg = T cos200, Horizontally: T sin200 = ma. a = g tan200 = 3.64 m/s2
Question 5
Car is either increasing speed to the West, or slowing down to the East. A, B, F
Question 6
(a) Greatest gradient is between 0 and 5 seconds
Question 7
 30 m
Convert speeds to m/s, then estimate area under graph up to 5 seconds
Question 8
a)
Acceleration = gradient of v-t graph = v/t = (7-1)/(5-1) = 1.5 ms-2
b)
c)
C
Distance travelled = area under v-t graph = average velocity x time
uv
1 7
)t  (
) x 4  16 m
=(
2
2
Fnet = ma = 60 x 1.5 = 90.0 N
Question 9
a)
v = u +at
o = 9.0 + 1.5a
a = -6.0 ms-2
Fnet = ma = 60 x (–6.0) = -360N (negative indicates retardation)
b)
J = p = Ft = -360 x 1.5 = -540 Ns
Question 10
a)
Ek = ½ mv2 = ½ x 60 x (3.0)2 = 270J
b)
Ugrav = mgh = 60 x 10 x 1.5 = 900J
c)
let F = average backward resistance force on boy and board.
Loss in Ugrav = increase in Ek + work done by frictional force
900 = 270 + 70F
630 = 70F
F = 9.0 N
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SOLUTIONS
UNIT 4.
PART F: Unit2 Revision
SOLUTIONS
Question 11
a)
If bag moves at constant velocity, the forward force = 20 N
W = Fs = 20 x 10 =200 J
b)
let the force in the direction of the cord = FN
then the horizontal component Fcos370 = 20
F = 20/cos370 = 25N
Question 12
Impulse = Area under the graph = 0.5 x 8000 x 0.8 = 3200 Joules = Mass x Change in Vel.
Vel = 3200/750 =4.26
4.3 m/s
Question 13
Unit of k = Unit of Force / (unit of vel)2 = kg.m.sec-2. m-2. sec2 = kg.m-1.
A
Question 14
Net Force = ma = (60 + 8) x 1.5 = 102 N = 1.0 x 102 N
Question 15
Net force = Friction - Air resistance
=> Friction = 102 + 5.5 = 107.5 N = 1.1 x 102 N
Consequential answer: Ans 65 + 5.5
Question 16
See Figure 1(Question 65). Two possible answers:
The two forces Fv, the vertical component of the force of the road on the wheel and FH, the horizontal
component of the force of the road on the wheel,
OR
the single force FR, which represents the total force of the road on the wheel. It, of course, must be
forwards of vertical.
Question 17
Distance = v x t = 19.4 x 0.750 = 14.6 m
Question 18
6.8 (or 6.2) ms-2s = 44 - 14.6 = 29.4 m u = 19.4 ms-1
v=0
a=?
2
2
v = u + 2as
=>
0 = 19.42 + 2a x 29.4
=>
a = 6.75 ms-2
[If s is taken as 45 m from the graph, a = 6.19 ms-2]
Consequential answer: 188.18 1 (44 – Ans17) or 188.18 / (45 – Ans17)
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UNIT 4.
PART F: Unit2 Revision
SOLUTIONS
Question 19
70 km hr-1 = ms-1 = 19.4 ms-1
Car's displacement (s) = vt = 19.4 x 8 = 156 m
Question 20
s = 140 m, u = 19.4 ms-1,v = 0
v2 = u2 + 2as
70 2
) + 2 x a x 140
36
a = -1.3 ms-2
0= (
Question 21
v = u + at
2
70 2
0 = ( ) - 1.35t
36
t = 19.44/1.35 = 14.4 s
Question 22
Frank requires 14.4 s to stop but the lights take only 8.0 s to change. Therefore, Frank will arrive at the
traffic lights 6.4 s after they have turned red.
Question 23
Weight force = mg (in direction shown above).
Resolved component of weight force parallel to road = mg sin 300
= 5250 N  5.3 kN
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UNIT 4.
PART F: Unit2 Revision
0
SOLUTIONS
Resolved component of weight force perpendicular to road = mg.cos 30
= 9093 N  9.1 kN
Question 24
(a) Resolved component of weight force perpendicular to the road is the normal reaction force
(N) of the road on the car (Newton's 3rd Law)
Frictional force (F) = N = 0.6 x mg cos 300 = 0.6 x 1050 x 10 x 0.8660 = 5.5 x 103 N
Question 25
Total force down the slope which needs to be overcome for car to travel at constant speed up the
slope = mg sin + N = 1050 x 10 x ½ + 5.4 x 103 = 5.3 x 103 + 5.5 x 103
= 1.08 X 104 N up the road.
Question 26
Energy is conserved
Ep(gravity) = Ek
(top of slide) (bottom of slide)
=½ mv2 = ½ x 30 x (8.0)2 = 960 J
Question 27
EP(grav) = mgh
30 x 10 h = 960
h = 960 / 300 = 3.2 m
Question 28
Energy = Power x Time = 5000 x 10 x 60 = 3.0 x 106J
Question 29
Energy =mgh = 60 x 10 x 1000 = 6.0 x 105 J
Question 30
Work done by force of boat on water = Force by distance. The force is 1000 N.
Work done = 1000 x 400 = 4.00 x 105 Joules
Question 31
Vel is constant, accel’n is zero, so net force is zero.
Question 32
Net force = Driving force - Water resistance. At 400 metres, water resistance = 1000 N, so driving force =
1000 N. At 200
m, water resistance = 500 N, so Net force = 500 N. Net force = mass x accel’n. So, accel’n = 500/800 =
0.62m/s2
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