6.3 – Work | Physics with Mr Shone

Definition: Work Done
Work done by an object is the amount of energy that has been converted from one form to another or transferred from one body to another.

Understanding
Being asked to “find the work done” is the same as being asked to “find the energy change“.

SI unit is joule (J).

Work is a scalar quantity.

Work is done when a force F acting on an object moved it a distance, d, in the direction of the force as given by

Equation
work done = force x distance moved in the direction of the force work = F × d Where work is the work done measured in joules (J) F is the force applied measured in newtons (N) d is the distance moved in the direction of the force measured in metres (m)

Equation

work done = force x distance moved in the direction of the force

work = F × d

Where

work is the work done measured in joules (J)

F is the force applied measured in newtons (N)

d is the distance moved in the direction of the force measured in metres (m)

Work Done by an object or on an object
When work is done by an object the object is losing energy. When work is done on an object the object is gaining energy.

Work Done by an object or on an object

When work is done by an object the object is losing energy.

When work is done on an object the object is gaining energy.

Is Work Done?
Is work done in lifting a child up into the air? Yes, we are doing work against gravity. The child ends up with more g.p.e. Is work done by person skiing down a mountain? Yes, work is done by gravity. The skier loses g.p.e as they move down the mountain. Is work done by a man pushing against a wall? No. There is no change of energy. The wall doesn’t move. Is work done by a mother pushing a baby cart? No. If the cart is moving at a constant velocity, along a level surface, then there is no change in k.e. or g.p.e.

Is Work Done?

Is work done in lifting a child up into the air?

Yes, we are doing work against gravity. The child ends up with more g.p.e.

Is work done by person skiing down a mountain?

Yes, work is done by gravity. The skier loses g.p.e as they move down the mountain.

Is work done by a man pushing against a wall?

No. There is no change of energy. The wall doesn’t move.

Is work done by a mother pushing a baby cart?

No. If the cart is moving at a constant velocity, along a level surface, then there is no change in k.e. or g.p.e.

Example

Find the work done when a force of 50 N moves an object through a distance of 40 cm in the direction of the force.

Work done = F × d

= 50 N × 0.40 m

= 20 J

A common mistake here may have been to forget to convert the length to metres.

Example
An object under the action of a force of 10 N, moves 2.0 m. What is the work done by the 10 N force on the object in each of the following scenario? (a) F is parallel to the direction of motion Work done = F × d = 10 N x 2.0 m = 20 J (b) F is perpendicular to the direction of motion Work done = 0 J (c) F is30° to the direction of motion Work done = F × d = (10 cos 30^o) (2.0) = 17 J

Example

An object under the action of a force of 10 N, moves 2.0 m. What is the work done by the 10 N force on the object in each of the following scenario?

(a) F is parallel to the direction of motion

Work done = F × d

= 10 N x 2.0 m

= 20 J

(b) F is perpendicular to the direction of motion

Work done = 0 J

(c) F is30° to the direction of motion

Work done = F × d

= (10 cos 30^o) (2.0)

= 17 J

Example
A man pulls a bucket of water slowly up from a well. Given that the mass of the bucket of water is 3.2 kg, calculate the minimum work done by the man to bring the bucket of water up from a depth of 5.0 m. Mass of 3.2 kg gives a weight of mg = 3.2 × 10 = 32 N To lift the bucket, we require a minimum force equal to the weight, T = 32 N Minimum work done = Force x distance = 32 × 5.0 = 160 J OR Minimum work done = energy converted to g.p.e = mgh = 3.2 × 10 × 5.0 = 160 J

Example

A man pulls a bucket of water slowly up from a well. Given that the mass of the bucket of water is 3.2 kg, calculate the minimum work done by the man to bring the bucket of water up from a depth of 5.0 m.

Mass of 3.2 kg gives a weight of mg = 3.2 × 10 = 32 N

To lift the bucket, we require a minimum force equal to the weight, T = 32 N

Minimum work done = Force x distance

= 32 × 5.0

= 160 J

Minimum work done = energy converted to g.p.e

= mgh = 3.2 × 10 × 5.0

= 160 J

Example

A block of weight 5000 N is at rest at the bottom of a slope. It is pulled up the slope for 50.0 m by an applied force of 1800 N. The friction between the block and the slope is 1 000 N.

(a) Determine:

(i) the work done by the pulling force,

Work done = F × d

= 1800 × 50.0

= 90 000 J

=90 kJ

(ii) the work done against friction,

Work done = F × d

= 1000 × 50.0

=50 000 J

= 50 kJ

(iii) the work done against gravity,

Work done = F × d

= 5000 × 3.0

= 15 000 J

= 15 kJ

(iv) the potential energy gained by the block.

EP = mgh

= 5000 × 3.0

= 15000 J

= 15 kJ

Note: WD against gravity = GPE gained by the object
It is not coincidence that the answers to (iii) an d (iv) are the same. They are in fact finding the same thing – The work done in lifting the object up (against gravity) is equal to the potential energy gained by the object .

(b) Using your answers to (a), determine the change in kinetic energy of the block as is it reaches the top of the slope.

Initial energy of the block is 0 J ( 0 J of g.p.e and 0 J of k.e.)

Energy given to the block is 90 kJ

Energy used by the block to overcome friction is 50 kJ

Thus energy possessed by the block at the end can be determined:

= 0 J + 90 kJ − 50 kJ = 40 kJ

As we know the block possess 15 kJ of g.p.e., the remaining 25 kJ must be k.e.

(c) calculate the speed of the block at the top of the slope.

As the k.e. is known to be 40 kJ, we can find the speed from

k.e. = ½mv²

25 000 = ½ × 500 × v²

v = 10 m/s

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