Transcript Lesson 2 RF equals ma Single objects

LESSON OBJECTIVE
Practise using RF = ma for single objects
Link to suvat equations
Question 1
A concrete block of mass 50 kg is to be lifted up the side of a building 50m tall.
If the block starts from rest and accelerates at a constant rate to reach a velocity of 1ms-1 at the
top of the building, find the acceleration of the block and the force in the rope required to produce
it.
Newton’s second law:
A body that is not in equilibrium must be accelerating/decelerating
The Resultant Force causing the acceleration is found using the
formula:
Resultant Force = Mass of the body × Acceleration
OR
Forces in Direction of Motion – Forces in Opposite Direction =
Mass of the body × Acceleration
Question 2
A van of mass 3000 kg is travelling at 10 ms-1 down a hill at an angle of 10o to the horizontal.
It must stop at the traffic lights at the bottom of the hill, a distance of 100m away.
If the brakes are applied to create a constant deceleration down the hill, find the force barking
force required.
100m
10o
Question 2
A van of mass 3000 kg is travelling at 10 ms-1 down a hill at an angle of 30o to the horizontal.
It must stop at the traffic lights at the bottom of the hill, a distance of 100m away.
If the brakes are applied to create a constant deceleration down the hill, find the force barking
force required.
3000gsin10 – B = 3000a
5105.26 – B = 3000a
B
N
100m
Need ‘a’
Constant deceleration
So use suvat u = 10 s = 100 v = 0
v = u2 + 2as
0 = 100 + 200a
a = -0.5
10o
Hence 5105.26 – B = -1500
B = 5105.26 + 1500
B = 6605.26 N
1) A stone of mass 50 grams is dropped into some liquid and falls vertically through it with
an acceleration of 5.8ms-2. Find the force of resistance acting on the stone.
2) A car of mass 700 kg is brought to rest in 7 seconds from a speed of 20 ms-1 by a
constant braking force. What force is necessary to produce this retardation?
3) A child on a sledge is being pulled up a smooth slope of 20o by a rope which makes an
angle of 40o with the slope. The mass of the child and sledge together is 20kg and the
tension in the rope is 170N.
Draw a diagram to show the forces acting on the child and sledge together.
a) Find the Normal Reaction Force
b) Find the acceleration of the sledge/child up the slope.
4) A small package P of mass 0.3 kg, which may be modelled as a particle, is placed on
the surface of a rough plane inclined at an angle θ to the horizontal, where sin θ = 3/5 ,
as shown. The friction force is 0.06g N. P is released from rest at a point 2 m from the
foot of the plane. Find
2m
P
a) the acceleration of P down the plane,
θ
b) the velocity with which P reaches the foot of the plane.
o
1) A stone of mass 50 grams is dropped into some liquid and falls vertically through it with
an acceleration of 5.8ms-2. Find the force of resistance acting on the stone.
0.05g – R = 0.05 x 5.8
R = 0.05g + 0.05x5.8 = 0.78 N
2) A car of mass 700 kg is brought to rest in 7 seconds from a speed of 20 ms-1 by a
constant braking force. What force is necessary to produce this retardation?
– B = 700 x a
u = 20
a=
v=0
t=7
So B = 2000N
v = u + at
0 = 20 + 7a
a= - 20/7
3) A child on a sledge is being pulled up a smooth slope of 20o by a rope which makes an
angle of 40o with the slope. The mass of the child and sledge together is 20kg and the
tension in the rope is 170N.
Draw a diagram to show the forces acting on the child and sledge together.
a) Find the Normal Reaction Force
b) Find the acceleration of the sledge/child up the slope.
170cos40 – 20gsin20 = 20a
a = 3.16
N
170
20o
20g
4) A small package P of mass 0.3 kg, which may be modelled as a particle, is placed on
the surface of a rough plane inclined at an angle θ to the horizontal, where sin θ = 3/5 ,
as shown. The friction force is 0.06g N. P is released from rest at a point 2 m from the
foot of the plane. Find
a) the acceleration of P down the plane,
b) the velocity with which P reaches the foot of the plane.
N
0.3g x 3/5 - 0.06g = 0.3a
0.06g
2m
P
a = 1.176/0.3 = 3.92
θo
0.3g
v2 = u2 + 2as
= 0 + 2x3.92x2
= 15.68
So v = 3.96 ms-1
1) A stone of mass 50 grams is dropped into some liquid and falls vertically through it with
an acceleration of 5.8ms-2. Find the force of resistance acting on the stone.
2) A car of mass 700 kg is brought to rest in 7 seconds from a speed of 20 ms-1.
What constant force is necessary to produce this retardation?
3) A child on a sledge is being pulled up a smooth slope of 20o by a rope which makes an
angle of 40o with the slope. The mass of the child and sledge together is 20kg and the
tension in the rope is 170N.
Draw a diagram to show the forces acting on the child and sledge together.
a) Find the Normal Reaction Force
b) Find the acceleration of the sledge/child up the slope.
4) A small package P of mass 0.3 kg, which may be modelled as a particle, is placed on
the surface of a rough plane inclined at an angle θ to the horizontal, where sin θ = 3/5 ,
as shown. The friction force is 0.06g N. P is released from rest at a point 2 m from the
foot of the plane. Find
2m
P
a) the acceleration of P down the plane,
θ
b) the velocity with which P reaches the foot of the plane.
o
A skier is being pulled up a smooth 25o dry ski slope by a rope which makes an
angle of 35o with the horizontal. The mass of the skier is 75kg and the tension in
the rope is 350 N. Initially the skier is at rest at the bottom of the slope. The slope
is smooth. Find the skier’s speed after 5 seconds and the distance they have
travelled in this time
35o
25o
N
350cos 25 – 75gsin25 = 75a
350
a = 0.454
25o
Constant acceleration
u=0
a = 0.454
t=5
v = u + at
v = 2.27
s = ut + ½ at2 = 5.68 m
75g
35o
Sam and his sister are sledging, but Sam wants to ride by himself. His sister gives him a push at
the top of a smooth, straight 15o slope and lets go when he is moving at 2 ms-1. He continues to
slide for 5 seconds before using his feet to produce a braking force of 95 N parallel to the slope.
This brings him to rest. Sam and his sledge have a combined mass of 30kg.
How far does Sam travel altogether?
Two sections:
Sliding freely
30gsin15 = 30a
N
Hence a = 2.54
Now s = ut + 0.5at2
30g
s = 41.75m in 5 seconds, final speed is 14.7 ms-1
N
F
Braking
30gsin15 – F = 30a
Hence a = -0.63
Now v2 = u2 + 2as
u = 14.7, v = 0, a =-0.63
Total distance travelled = 14.7 + 171.5 = 213m
30g
s = 171.5
Hints:
Remember the
relationship between cos
x and sin (90 - x)
Do what you would
normally do.
What does M sliding
mean about the
acceleration of M?
How do you find max and
mins?
Express the whole thing in
terms of tan
Put a pair of scales in a life and stand on them.
What do the scales read?
What happens when the lift starts moving?
Fred Lemming has a mass of 10kg.
He stands in a lift.
Three situations:
Lift is stationary
Find the
Tension in the
cable and the
normal contact
force
Lift accelerating
upwards
Find the
Tension in the
cable and the
normal contact
force
Lift deccelerating
upwards
Find the
Tension in the
cable and the
normal contact
force