WORK- Is the amount of force applied to an object over a distance

3 CONDITIONS FOR WORK TO BE DONE

- There must be a force acting on the object.

- The object has to move a certain distance called the displacement.

- There must be a component of the force in the direction of the motion.

WORK IS NOT DONE WHEN :

- The object is stationary

- No force applied on the object in the direction of the motion

- The direction of the motion of the object is perpendicular to that of the applied force

- When work is done and energy is transferred to the object

A B

A B

A boy applies a force to a wall and becomes exhausted.

Fd

A waiter carries a tray full of meals by his arm across the room

Fd

CALCULATING THE AMOUNT OF WORK

( F cos θ ) dW =

d – displacement | meters (m)

F – force parallel to the displacement | newton (N)

Θ – the angel between the force and the displacement

W – work | newton-meters or joules ( J )

A book weighing 1.0 N is lifted 2 m. How much work was done?

It took 50 J to push a chair 5 m across the floor. With what force was the chair pushed?

A force of 100 N was necessary to lift a rock. A total of 150 J of work was done. How far was the rock lifted?

- 2 J

- 10 N

- 0.666 m

DID YOU KNOW ?

If you have to lift a new sofa to a second floor apartment, the work done against the gravity is the same whether you haul it straight up the side of the building with ropes or take a longer path up the stairs.

Only the vertical distance matters because the force of gravity is vertical

A box is dragged across a

floor by a 100N force directed

60o above the horizontal. How

much work does the force do in

pulling the object 8m?

Try to solve:

A porter pulls a 10kg luggage along a

level road for 5 m by exerting a force of 20N

at an angle of 30o with the horizontal

shoulder through a vertical distance of 1.5

meters and carries it for another 5 meters.

How much work does he do in

a.) pulling

b.) lifting

c.) carrying the luggage on his shoulder

Solve:

a.) pulling the luggage

Given: F – 20 N Find: WΘ – 30o

d – 5 m

Sol’n : W = ( F cos θ ) d= (20 N) (cos 30o) (5m)

W = 87 J

b.) lifting the luggage

Given : m – 10 kg Find: W

d – 1.5 m

Sol’n : W = F x d

F = ?

= (98 N) (1.5m)

W = 147 J

The force on the luggage is perpendicular to the direction of motion. The distance moved in the direction of force is zero. Therefore, work is zero.

Hence, the porter does no work in carrying the luggage.

c.) carrying the luggage

W = O

ENERGY- The capacity to do work

- the energy possessed by bodies in motion

Kinetic Energy

Potential Energy

-stored energy

- Associated with forces that depend on the position or configuration of a body and its surroundings.

POTENTIAL ENERGY: THE STORED ENERGY

|Gravitational Potential Energy|

- the energy of an object due to its higher position in the gravitational field

PEg = mgh

Solve :

A 800g ball is pulled up a slope

as shown in the diagram. Calculate

the potential energy it gains.

20 cm50 cm

A box has a mass of 5.8kg.

The box is lifted from the garage

floor and placed on a shelf. If the

box gains 145J, how high is the

shelf?

Solve :

|Elastic Potential Energy|

- stored energy in a spring by stretching or compressing it

Pes = 1

2𝑘𝑥2

k – spring constant

x – stretched distance of the elastic object

k xF

When a 4 kg mass is hung vertically on a

certain light spring that obeys Hooke's

law, the spring stretches 2.5 cm. If the 4

kg mass is removed,

(a)how far will the spring stretch if a 1.5

kg mass is hung on it, and

(b)(b) how much work must an external

agent do to stretch the same spring 4

cm from its unstretched position?

(a)We find the spring constant of

the spring from the given data.

F = kx.

F = mg = -(4 kg)(9.8 m/s2) = -39.2N.

k = F/x = (39.2 N)/(0.025 m) = 1568

N/m.

Now we use x = F/k to find the

displacement of a 1.5 kg mass.

F = (1.5kg)(9.8m.s2)

= 14.7 N.

x = (14.7 N)/(1568 N/m)

= 0.009375 m

= 0.975 cm.

(b) W = (1/2)kx2

= (1/2)(1568 N/m)(0.04m)2

= 1.2544 Nm

= 1.2544 J

A 5 cm stretched spring with a 15 g

bag of salt hanging on it is lifted up

by 7 meters.

Calculate its elastic potential energy .

KINETIC ENERGY: ENERGY IN MOTION

KE = 1

2𝑚𝑣2

m - mass of the object

v – velocity

Solve :

calculate the kinetic energy of

a 1000 kg car travelling at 60 km/h.

KE = 1

2𝑚𝑣2

= 1

2(1000kg)(60𝑘𝑚/ℎ)2

= 138 944.45 J

A van has a mass of 3000 kg and

a car has a mass of 1500 kg.

if both are travelling at the same

speed, how would you compare

their kinetic energy?

Solve :

Van : KE = ½ (3000 kg) 𝑣2

KE = 1500 𝑣2

Car : KE = ½ (1500 kg) 𝑣2

KE = 750 𝑣2

The van’s KE is twice the KE of the car.

POWER- the rate of doing work

CALCULATING POWER

P = 𝑊

𝑡W = work done | newton meters (J)

t = time | seconds (s)

P = power | joules per second (J/s)

or watts (W)

Solve :

Dan climbs a flight of stairs in

1.5 minutes. If he weighs 450 N and

the stairs is 10 meters from the

ground, how much power will he

develop?

Given : t = 1.5 min / 90 sF = 450 Nd = 10 m

P = 𝑊

𝑡

= (450 𝑁)(10𝑚)

90 𝑠

= 50 N ● m/s

or 50 W

DID YOU KNOW ?

The unit of power, watt (W),

was named after the Scottish mathematician and engineer –

JAMES WATT

Solve :

Andy and Bryan each lift a 150 kg barbell at

a height of 1.5 m off the ground. Andy lifts

his barbell in 1 second and Bryan lifts his

barbell in 2 seconds.

a.) who does more work ? Explain.

b.) who exerts more power ?

a.) both do the same amount of work.

b.) Andy : P = W/t

P = (150 kg) (9.8 m/s2) (1.5 m)

1 sP ≈ 2200 watt

Bryan : P = W/t

P = (150 kg) (9.8 m/s2) (1.5 m)

2 s

P ≈ 1100 watt

SIMPLE MACHINES