ding Thethe Effects of a Force

8/6/2019 ding Thethe Effects of a Force

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2.5 UNDERSTANDING THE EFFECTS OF A FORCE

FORCE

Definition :Anything that changesthe state of rest ormotion of a body movingin a straight line

Effects :*change the size of the

body*change the shape of the

body*change the resting state of

the body.* change the state of the

body* change the direction of

Vector quantity

S.I units :

Kgms-2 or Newton (N)

unbalance when there are

a net force acting on it.

Balance if two or more external forcesacting on it cancel out one another andthe net force of resultant force are zero

The net forceacting on the

books is zero.R-W = 0R is a normalreactionW is weight.

The book remainat rest.

The net force acting onthe car is zero.

F-T = 0F is a frictional forceT is driving force

provides by the engine

The car moving atconstant velocity along astraight road.

When four forces are balancethe net force acting on the

plane is zero.L-W=0 therefore L=WT-F=0 therefore T=F

An airplane is flyinghorizontally at a constantheight with a constantvelocity.


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Newton Second Law of Motion

The rate of change of momentum is directly proportional to the resultant force and inthe same direction as the force.

F rate of change of momentumF (m mu)

tF m (v u) but v u = ma

t t

F maF = kma where kis a constant Fis a force in newtons (N)

m is the mass in kilograms (kg)a is the acceleration in metres per second,

(ms-2)

1 N of force is defined as the magnitude of a force required to accelerate 1 kg of a

body by 1 ms-2. Hence k= 1

Therefore

The relationship between F, m and a can be shown by this graph :

a a

a varies inversely with m a is proportional to 1 a varies directlym with F

F = ma

m

F

1

m

a


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Example 1 :

The block of wood of mass 8 kg is pulled on the floor by a force of 12 N and having aconstant velocity . What is the acceleration, if the block of wood is pulled by a 60 Nforce ?

solution :

constant velocity means that the net force acting on this object is zero. Thereforethe frictional force is also 12 N.

F = ma

60 -12 = 8 x aa = 48

8

a = 6 ms-2

Example 2 ;

A particle which is initially at rest is accelerated for 10 seconds. The force thatis applied upon the particle is 25 N. If the final velocity of the particle is 20 ms

-1

and that the frictional force between the particle and the surface is neglected,what is the mass of the particle ?

solution :information given : t= 10 s

F = 25 Nv= 20 ms-1u= 0 ms-1 ( initially at rest )

using formula F = m (v - u)t

25 = m (20 0 )10 therefore m = 12.5 kg

v constant

12 N 60 N12 N

v constant

12 N 60 N

frictionalforce 12 N


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2.6 Analysing Impulse and Impulsive Forces

Ways to Reduce Impulsive Force

1. Bending the legs to land on the ground when jumping.2. Using carpet3. The design of the car :

*using a soft metal in front and the rear part of the car, so that thecar easily crumpled during an accident the time for the car to crumpledincreaseso that the impulsive force will decrease.

* The seats of passengers are strengthened - protect the passengers.

* Safety belts slow down the motion of the passengers.* Air bags prolong the time of action and reduce the impulsive force.

4. Moving hand to the back while catching the ball decrease the impulsive forceand prolong the time of action.

5. Keeps fragile goods in a soft polystyrene- longer time taken to diffuse the bigforce and reduce the impulsive force.

Impulse

Definition :

Change of momentum

Ft= mv - mu

S.I unit :kg m s-1

orN s

Impulsive Force

Definition :rate of change of

momentum F= mv mut

S.I unit:kg m s-1

or

N

* When the time of action is prolonged, the impulsive forcewill decrease.

* When the time of action is shortened, the impulsive forcewill increase.


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2.7SAFETY FEATURES IN VEHICLE

A Good Use of

Impulsive Force

-Shorten the time- increase the

impulsive force

Hockey stick and golf club* made of hard wood and

hard alloy.

Martial art player is able tobreak a pile of brick easily* the hand move very fast

and stops when it hits thetop bricks.

The pile and the pilediver.* made of hard steel

alloy* release very fast to

hit the hard pile

The pestle and mortar.* made of hard material* the pestle moves very fast and th

mortar stops the motion of thepestle in a short time.


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2.8 UNDERSTANDING GRAVITY

Example :Do a feather and coin fall with the same acceleration in air ? Why ?

solution :

No, because the the coins has small surface area than that of the feathertherefore the air resistance which acts on the feather is higherand theacceleration of the coin is biggerthan that of the feather.

2.8 UNDERSTANDING GRAVITATIONAL FIELD

Example 1:

definition :Any object which falls only underthe influence of the force ofattraction due to gravity andwithout any influence of otherforces

fall with the same

acceleration due togravity.g = 10 ms-1

Example of free fall :

Both object reach thebottom at the sametime.

FREE FALL

Definition :the gravitational forceacting per unit mass onan object in the field

On Earthg= 9.8 N kg-1

can be found by using thefollowing equation :g = Gravitational force , F

Mass of body, m

GRAVITATIONAL

FIELD


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What happens to the value of the gravitational field strength when we move awayfrom the earth ?

solution :The gravitational field strength gets less when we go further away from the earth.

Example 2 :A body of mass 2 kg has a weight of 20 N . Find the value of the gravitationalfield strength .

solution :g = 20 N

2 kg= 10 N kg-1

Example 3 :An object has a mass of 8 kg. Find its weight on (a) the Earth (b) the moon.( The acceleration due to gravity on the Earth is 9.8 m s

-2while on the Moon is

1.6ms-2)

solution :

a) F = mg (b) F = mg= 8 kg x 9.8 m s

-2= 8 kg x 1.6 m s-2

= 78.4 N = 12.8 N


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ACTIVITY(50 MINUTES)

2.5 UNDERSTANDING THE EFFECTS OF A FORCE1.Complete the table below.

Forces acting on an object Resultant Force(FR)

(i)20 N 20N

FR = Balanced / Unbalanced

(ii)

60 N 100 NFR = Balanced / Unbalanced

(iii)8N 5N

10NFR = Balanced / Unbalanced

2.Complete the following table which shows the relationship between resultantforce, mass and acceleration.

Mass/kg Acceleration/ms-2

Resultant force/N

5 10

2 20

10 500

3. Solving problems using F = ma

(a) State Newtons Second Law of motion.

(b) A box of mass 3 kg is put on a smooth floor.The box is pushed horizontallywith a force as shown below.What is the acceleration of the box?

3N


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(c)F = 2.0N

A toy car of mass 800g is pulled along a level runway with a constant speed by aforce of 2N.(i) What is the friction on the toy car?(When a car moving with a constant speed, the resultant = 0N)

(ii) When the force is increased to 6 N, what is:

The unbalanced force acting on it?

The acceleration of the toy car?

2.6 ANALYSING IMPULSE AND IMPULSIVE FORCE.

4. A car with a mass of 1000kg is moving with a velocity of 25ms -1.It then collideswith a tree and is stopped in 0.05 s.Calculate the impulse and impulsive forceacted on the car.

Mass of the car =

Initial velocity of the car,u =

Final velocity of the carv=

Time taken to stop =

Therefore Impulse(Ft) = change of momentum(mv mu)

=

The impulsive force = rate of change of momentum

=


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2.7 SAFETY FEATURES IN VEHICLES

5. The picture below shows a crash test of a vehicle conducted by a carmanufacturer.

(a) State the labelled parts in the picture.

P

Q

R

S

(b) Describe the safety features of the following parts.

P

Q

R

S

(c) The major consideration to design safety features of a car is to reduce the

during an accident.The longer the ,the smaller will be the

impulsive force acting on the car.


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2.8 UNDERSTANDING GRAVITY

6. A spacecraft of mass 1200kg experiences a gravitational force of 6000N.Calculate the gravitational field strength at the location of the spacecraft.

Mass,m =

Gravitational Force,F =

Therefore Gravitational field strength,g = F/m

=

7.An astronaut has a mass of 75kg.what is his weight if

(a) he is on the surface of the Earth where the gravitational field strength is9.8Nkg-1?

w = m x g

(b) he is on the surface of the moon where the gravitational field strength is 1/6of that on the surface of the earth?

gmoon = 1/6x gearth

=

weight w = m x gmoon

=


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ASSESSMENT

OBJECTIVE QUESTIONS

1. Force is defined as

A the change in impulse

B the difference in pressure

C the rate of change in momentum

2. An object is moving with zeroresultant force.Which of the followinginference is not true?

A Its velocity is constant.

B Its acceleration is zero

C Its displacement is zero

D It has encountered a constant

resistance

3. An object of fixed mass will movewith a constant acceleration, whenthe resultant force acting on it is

A proportional to its displacement.

B constant but not zero.

C increasing.

D zero.

4. A truck has a mass of 5000 kg.what isthe braking force needed to slow itdown from 20ms-1 to a complete stopin 2.0 s

A 50 000 N D 500 N

B 5000 N E 50 N

C 1000 N

5 20N 20N

The above figure shows two woodenblocks each weighing 4kg are beingpulled by a force of 20N resulting inan acceleration of2ms-2.If one of theblocks is being pulled by the sameforce of 20N as shown above,whatwill the acceleration, in ms-2 be?A 2.0 D 4.5B 3.0 E 5.0C 4.0

6 The acceleration of free fall on themoon is 1.6ms-2.The acceleration of

free fall on the earth is 10ms

-2

. A rockhas a mass of 10kg on theearth.Which of the followingstatements of the rock is correct?

A Its mass on the moon is 16N.

B Its mass on the moon is 10kg.

C Its weight on the moon is zero.

D Its weight on the earth is 10N


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STRUCTURED QUESTION.

1. Figure1.0(a) shows a trolley P on a horizontal bench.Figure1.0(b) shows anidentical trolley Q loaded with steel bars on the same bench.A horizontal force of10.0N acting for a time of 0.2s is applied to each trolley in turn.

10N 10N

0.2s 0.2s

Figure 1.0(a) Figure 1.0 (b)(a) After 2.0 s ,trolley P moves to the right with a speed of 5.0 ms -1 but trolley

Q moves to the right with a speed of 1.0 ms-1

.Explain why these speedsare different.

(1 mark)(b) Trolley P is now turned overso that it is resting on its top surface as

shown in figure 1.1.

10 N0.2s

figure 1.1

Explain why the force of 10.0N acting for 0.2s now results in a very smallspeed

and why that speed rapidly reduces to 0 N.

(1 mark)(c) The mass of trolley P is 0.40 kg. determine the weight of trolley P when

the(i) trolley is nearthe surface of the earth

(1 mark)

(ii)the surface of the moon

(1 mark)Given that the acceleration of free fall is 10.0 ms

-2at the surface of the earth and

1.6 ms-2at the surface of the moon.

P P


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2. Figure 2.0 shows a raindrop falling.

B

AFigure 2.0

A is the force which causes the raindrop to fall, B is the force whichopposes the motion

of the raindrop.

(i) Name force A.

(1 mark)

(ii) Name a possible cause of force B.

(1 mark)

(iii) Explain what happens to the raindrop when force A equals to force B.

(2 marks)


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3.A man skydives from a stationary helicopter and parachutes down.Figure3.0 shows three positions during the fall.The drag force Ron the parachutistand the total weightof the parachutist Win each position are shown in thediagrams.

Figure 3.0

(a)At which position is he falling with constant velocity?Give reason for youranswer.

(2 marks)(b)At which position is he falling with an upward acceleration. Find the

acceleration at that instant.

(2 marks)(c )At which position is he falling with a downward acceleration? Find the

acceleration at that instant.

(2 marks)


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ANSWER KEY.

(Activity)

1(i) 0N, Balanced , (ii) 40 N, to the right(Unbalanced), (iii) 13 N,to the left(Unbalanced)

2. 50N, 10kg, 50ms-2

3.(a) The acceleration of a body, a , is directly proportional to the net force actingon it,F, and inversely proportional to its mass,m:

F = ma

(b) F = ma

a = F/m

= 3/3

= 1 ms-2

(c) (i)Friction Force = Force applied = 2.0 N

(ii) The unstable force = 6 N 2 N

= 4 N

The acceleration a = F/m

= 4/0.8

= 5 ms-2

4. m = 1000 kg , u = 25ms-1

,v = 0ms-1

, t = 0.05 s

Therefore impulse = 0 25000 = -25000Nm

Impulsive Force = -25000 / 0.05

= -500 000 N


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5(a) Parts of a car.

P Rubber front bumper.

Q Front crumple zone.

R Windscreen.

S Air bag

(b) Safety features.

P Absorbs impact to prevent damage to the car.

Q Easily crushed to increase the time interval of impact.

R Made of shatter-proof glass to reduce injuries caused by the pieces ofglass which scatter during the collision.

S Cushion the passengers from direct impact with the steeringwheel/windscreen.

( c ) time/time taken.

6. m = 1200 kg

F = 6000N

g = 5.0 Nkg-1

7. (a) w = 735 N

(b) w = 122.5 N


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ANSWER KEY.(Assesment)

Objective questions.

1. C2. C3. B4. A5. D6. B

Structured Question.

1.(a) The total mass of the steel bars and the trolley Q is greater than the massof trolley P.Since the same force is applied in both cases,the acceleration oftrolley Qis smaller than the acceleration of trolley P.

(b) When the trolley P is overturned, the frictional force between the trolley and

the bench is greater.

(c ) (i) Weight near surface of the earth = 0.4 x 10= 4.0 N

(ii) Weight near surface of the Moon = 0.4 x 1.6= 0.64 N

2. (a) weight or gravitational force.

(b) Air resistance

(c ) When A = b ,by Newtons Second Law,

resultant forceF = ma becomes

A B = maTherefore a = A B

m= 0 ms-2

The raindrop moves at constant velocity.

3. (a) Constant velocity at Z. At Z resultant force = 0 N.

(b)Upward acceleration at Y. At Y, resultant force is 400N upwards.

(c) Downward acceleration at X.At X, resultant force is 800N downwards.

ding Thethe Effects of a Force

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