1. Forces 2nd Law

2. Starter: KUS objectives BAT Understand Newtons 3 laws
BAT Resolve forces for moving objects in a straight line
BAT Solve problems using Newtons 2nd law
Starter:
A car accelerates constantly from 22 ms-1 to 48 ms-1 over 13 seconds. Find the distance travelled in this time
𝑠= 𝑢+𝑣 2 𝑡 = 70 2 ×13 =455 𝑚
A cannonball is fired from a cannon and travels 1048 m in 4 seconds before hitting the ground. What was its initial velocity?
Vertically 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 =0
0= 𝑢 −
𝑢= 19,6 𝑚𝑠 −1
Horizontally 𝑠=𝑢𝑡 =1048
1048=4𝑢
𝑢= 𝑚𝑠 −1
Initial velocity 𝑢= = 𝑚𝑠 −1
at an angle above horizontal of 𝑡𝑎𝑛 − =4.28° 2

3. Newton’s Laws:
1st law. A particle will remain at rest or will continue to move with constant velocity in a straight line unless acted on by a resultant force
2nd law. The force applied to a particle is proportional to the mass m of the particle and the acceleration produced
3rd law. Every action has an equal and opposite reaction
F = ma
For moving objects the second law gives us
We already met the second law with Weight = mg where g = 9.8 ms-2
The third law is useful where we have normal contact and thrust forces
What might the opposing forces be called? 3

4. 7i + 13j 4i + 4j -2i - 5j The forces remain constant
WB1 Stone
A stone of mass 600 g is dragged by forces parallel to the ground
If the forces are (7i + 13j) N, (4i +4j) N and (-2i – 5j) act on it.
Suggest a suitable model
7i + 13j
4i + 4j
Plan view
-2i - 5j
What might the opposing forces be called?
The forces remain constant
There are no other forces
The stone is modelled as a particle

5. A stone of mass 600 g is dragged by forces parallel to the ground
WB1 solution
A stone of mass 600 g is dragged by forces parallel to the ground
Find the acceleration of the stone when forces of (7i + 13j) N,
(4i +4j) N and (-2i – 5j) act on it. Find also the magnitude and direction of the acceleration
4i + 4j
7i + 13j
-2i - 5j
The resultant force is
−2 −5 =
= 9i + 12j
Now use F = ma
9i j = 0.6 a
a = (15i + 20 j) ms-2
What might the opposing forces be called?
The Magnitude of acceleration = = 25 ms-2
And the direction of a is given by tan − = 53.1o 5

6. T R 300 W The Weight is constant, i.e. g is constant
WB2
A Sledge of mass 80 kg is being dragged along the ground by a rope at an angle of 30, such that it is accelerating horizontally.
Suggest a model T R
300 W
The Weight is constant, i.e. g is constant
The rope is inelastic / inextensible
The angle and size of Tension stay constant
There are no resistant forces
The sledge is modelled as a particle

7. WB2
A Sledge of mass 80 kg is being dragged by a force of 200 Newton's at an angle of 300 to the horizontal smooth surface such that it is accelerating horizontally forward as shown.
Find the distance travelled from rest in 5 seconds
200 cos sin 30
0 𝑅
200
The sledge is ‘balanced’ so that forces up = forces down
R = 784
300
0 −784
Explain that the sledge is NOT moving up or down – only horizontally forward
HORIZONTALLY F=ma cos30 = 80 a
𝒂 = 𝟐.𝟏𝟕 𝒎𝒔 −𝟐
now using 𝒔 = 𝒖𝒕+ 𝟏 𝟐 𝒂 𝒕 𝟐 = 𝟎 + 𝟏 𝟐 𝟐.𝟏𝟕 𝟐𝟓 =𝟐𝟕.𝟏 𝒎

8. T R Fr 300 W The Weight is constant,
WB3
The same Sledge is now dragged by the same force along a rough surface such that resistance force is 80 Newton’s.
It is still accelerating horizontally forward
Suggest a model T R Fr
300 W
The Weight is constant,
Resistance force (friction) is constant
Rope is inelastic / inextensible
The angle and size of Tension stay constant
The sledge is modelled as a particle

9. WB3 solution
The same Sledge (mass 80 kg) is now dragged by the same force along a rough surface such that resistance force is 80 Newton’s
If the sledge initially had speed 5 ms-1, how far has it travelled when it reaches a speed of 15 ms-1
300
200 cos sin 30
0 𝑅
0 784
80 0
HORIZONTALLY F=ma 𝟐𝟎𝟎cos𝟑𝟎 −𝟖𝟎 = 𝟖𝟎 𝒂
𝒂 = 𝟏.𝟏𝟕 𝒎𝒔 −𝟐
now using 𝒗 𝟐 = 𝒖 𝟐 +𝟐𝒂𝒔
𝟏𝟓 𝟐 = 𝟓 𝟐 +𝟐(𝟏.𝟏𝟕)𝒔
𝒔 = 𝟖𝟓.𝟓 𝒎

10. Find the tension in the string
WB4 A Particle of mass 2 kg is attached to the lower end of a string hanging vertically. The particle is lowered and moves with acceleration 0.2 ms-2
Find the tension in the string T W
a = 0.2
Draw a diagram:
2nd Law
What might the opposing forces be called? 10

11. The child exerts a force on the lift so there is an
WB4 A lift is accelerating upward at 1.5 ms-2. A child of mass 30 Kg is standing in the lift. Treating the child as a particle find the force between the child and the floor of the lift. R W
a = 1.5
The child exerts a force
on the lift so there is an
opposite force R upwards in the direction of movement
Draw a diagram:
2nd Law
What might the opposing forces be called? 11

12. WB5a
A parcel of mass 5 Kg is released from rest on a rough ramp of inclination
θ = 300 and slides down the ramp. The resistance due to friction is 8 N
Treating the parcel as a particle, find the acceleration of the parcel 5g R Fr
300 W
Draw a diagram:
W = −5𝑔 sin 30 −5𝑔 cos = −24.5 −42.4 12

13. 𝐹= 𝑚𝑎 𝑊−𝐹𝑟= 𝑚𝑎 24.5−8= 5𝑎 𝑎=3.3 𝑚𝑠 −2 R Fr 5g W
WB5b A parcel of mass 5 Kg is released from rest on a rough ramp of inclination
θ = 300 and slides down the ramp. The resistance due to friction is 8 N
Treating the parcel as a particle, find the acceleration of the parcel 5g R Fr W
we want to look at this resultant force in the direction of slope only
W = −24.5 −42.4
𝐹= 𝑚𝑎
2nd Law
𝑊−𝐹𝑟= 𝑚𝑎
24.5−8= 5𝑎
𝑎=3.3 𝑚𝑠 −2 13

14. W R D R = 2265.8 Resultant Force up slope = ma F = ma
WB6 A car with a constant driving force of N meets a slope inclined at an angle of elevation of The resistance forces to the car on the slope are 400 N.
If the car has weight 2500 N, find the acceleration of the car up the slope W R D
R =
Resultant Force up slope = ma
F = ma
1500 −400 −1056.5= 𝑔 𝑎
43.5 = 2500 𝑔 𝑎
W = −2500 sin 25 −2500 cos = −
𝑎=0.17 14

15. WB7
A parcel of mass 3 kg is sliding down a smooth inclined plane with an acceleration of 4 ms-2 . Find the angle of inclination of the plane 3g R
R = 3𝑔 cos 𝜃
Resultant Force up slope = ma
F = ma
3𝑔 sin 𝜃 = 3 ×4
sin 𝜃 = 12 3𝑔 =0.408
W = −3𝑔 sin 𝜃 −3𝑔 cos 𝜃
𝜃=24°

16. WB8
Daisy is sledging down a slope of 300 and accelerating at 4 ms-2.
The resistance force due to friction is 10 N. Find Daisy’s mass. mg R Fr
R = 𝑚𝑔 cos 30
Resultant Force up slope = ma
F = ma
m𝑔 sin 30 −10 = 𝑚 ×4
m (𝑔 sin 30 −4) = 10
m = 10 𝑔 sin 30 −4 =11.11 𝑘𝑔
W = 𝑚𝑔 sin 30 𝑚𝑔 cos 30 16

17. One thing to improve is –
KUS objectives
Understand Newtons 3 laws
Solve problems using Newtons 2nd law
self-assess
One thing learned is –
One thing to improve is –

18. END