Lower Secondary Science

[gjooheng]

f(b) = 2 X 3.001^2 - 1/3.001 = 17.6788

f(a) = 2 X 3 ^2 - 1/3 = 17.6667

[f(b)-f(a)] / 0.001 = 12.1 (3 sig fig)

Compared with differentiation

dy/dx = 4x + 1/x^2
gradient at x= 3, = 109/9 = 12.1 (3 sig fig)

For Emaths students who have not learnt differentiation, the calculator approach is an alternative way to check the gradient.

[Verysuperkiasu]

Hi ....


My dd is rather confused about work done. It says in her text book that work done on an object is force exerted on the object multiply by the distance moved by the object in the same direction.

Case study :

A student lifts a box weighing 50N thru a vertical height 1.1m and then walks horizontally for 2 m while holding the box. What is the work done by the student?

Workdone on the box being lifted = F x d
= 50N x 1.1m = 55J

As for horizontal movement, according to the text bk, the horizontal force is zero. So work done horizontally is also zero.

My question: The fact that the student pushes himself forward to overcome frictional force when walking means that there is force and therefore work is done horizontally. Unless the text book means horizontally no work is done on the box (which they did not make clear)???

Can I clarify that horizontally there is still work done by the walking motion but this has no bearing on the work done on the box?

Also, I know the unit for workdone is Joules. But is it wrong to put the unit as Nm?

[Dr.033430Daniel]

Verysuperkiasu:

Hi ....


My dd is rather confused about work done. It says in her text book that work done on an object is force exerted on the object multiply by the distance moved by the object in the same direction.

Case study :

A student lifts a box weighing 50N thru a vertical height 1.1m and then walks horizontally for 2 m while holding the box. What is the work done by the student?

Workdone on the box being lifted = F x d
= 50N x 1.1m = 55J

As for horizontal movement, according to the text bk, the horizontal force is zero. So work done horizontally is also zero.

My question: The fact that the student pushes himself forward to overcome frictional force when walking means that there is force and therefore work is done horizontally. Unless the text book means horizontally no work is done on the box (which they did not make clear)???

Can I clarify that horizontally there is still work done by the walking motion but this has no bearing on the work done on the box?

Also, I know the unit for workdone is Joules. But is it wrong to put the unit as Nm?

For your last question, it is better to express energy as Joules. As a side note, the unit of moment is Nm, which is not written as Joules since in the case of moments, the force is perpendicular to the distance from the fulcrum.

I agree with you that question could be phrased better. It's clearer to talk about work done on the box. If the student is walking horizontally at constant velocity, then we could argue there is no net force on the box in the horizontal direction. Just thinking about the box itself, it is moving at a constant velocity horizontally. If air friction is negligible, then you could say that horizontally the student does not need to exert a force on the box so long as he is walking at constant velocity as he lifts the box.

The main point of the question is that a vertical lifting force cannot account for work in the horizontal direction. For work, force must be in the same direction as the motion.

Another other version of this problem is luggage on a conveyor belt being lifted up and moved horizontally into the luggage compartment of an airplane. Think of luggage on a slanted incline conveyor belt. That's another case where moving the luggage up requires work, but if it moves horizontally at constant velocity with no friction opposing the motion, then there is no work in the horizontal direction.

And yes you are right about the idea if you have to exert a force on yourself while walking to maintain a constant speed then yes you are doing work in the horizontal direction. The faster you walk or run the more air friction there will be and the more force and work you would need.

So as a student answering a structured question like this, you could write \"assuming no horizontal force on the box\"... and then write your solution.

[Verysuperkiasu]

Dr.Daniel:

For your last question, it is better to express energy as Joules. As a side note, the unit of moment is Nm, which is not written as Joules since in the case of moments, the force is perpendicular to the distance from the fulcrum.

I agree with you that question could be phrased better. It's clearer to talk about work done on the box. If the student is walking horizontally at constant velocity, then we could argue there is no net force on the box in the horizontal direction. Just thinking about the box itself, it is moving at a constant velocity horizontally. If air friction is negligible, then you could say that horizontally the student does not need to exert a force on the box so long as he is walking at constant velocity as he lifts the box.

The main point of the question is that a vertical lifting force cannot account for work in the horizontal direction. For work, force must be in the same direction as the motion.

Another other version of this problem is luggage on a conveyor belt being lifted up and moved horizontally into the luggage compartment of an airplane. Think of luggage on a slanted incline conveyor belt. That's another case where moving the luggage up requires work, but if it moves horizontally at constant velocity with no friction opposing the motion, then there is no work in the horizontal direction.

And yes you are right about the idea if you have to exert a force on yourself while walking to maintain a constant speed then yes you are doing work in the horizontal direction. The faster you walk or run the more air friction there will be and the more force and work you would need.

So as a student answering a structured question like this, you could write \"assuming no horizontal force on the box\"... and then write your solution.

Thank you very much, Dr Daniel....for the clarification.

[DJ123]

Hi, Dr Daniel

Do you have any recommendation for sec 2 sci assessment book.

[Dr.033430Daniel]

DJ123:

Hi, Dr Daniel

Do you have any recommendation for sec 2 sci assessment book.

Yes the book that I recommend is called \"Lower Secondary Science Conceptual Exercises\" author: G.S. Cheng. It is red in colour and the publisher is GLM.

This is good for many but not all students in the express stream. If the school has a challenging express syllabus, or if the student is on an IP or IB track, then my strategy is to make sure they master the level in this book first and then I move them on to Sec 3 level books and worksheets that I have.

[DJ123]

Thank you

[DJ123]

Thank you

[Verysuperkiasu]

Hi Dr Daniels…


Since the unit for energy is in Joules (for eg potential energy of a ball is 50 Joules) then is this energy also equals to the workdone to the ball? In which case you do not need to apply the formula Workdone=Force x Dist moved in the same direction?

[Dr.033430Daniel]

Verysuperkiasu:

Hi Dr Daniels..


Since the unit for energy is in Joules (for eg potential energy of a ball is 50 Joules) then is this energy also equals to the workdone to the ball? In which case you do not need to apply the formula Workdone=Force x Dist moved in the same direction?

I’ll answer with some specific numbers.

If a ball with a mass of 1kg has a gravitational potential energy of 50J, then using the formula for potential energy,
mgh = 50 J
(1kg) (10m/s2) h = 50 J

Solving for h the height is 5 meters. So this ball is 5 meters high.
Suppose a question asks how much work is needed to raise the ball to a level of 5m (assuming no air friction). The answer would be 50 Joules.

Putting this in a formula, the force needed to move the ball the distance of 5m would be a force equal to the weight of the ball. Force is mg.

F = mg = (1kg) (10m/s2) = 10N

W = 10N (5m) = 50J

A force of 10N acting over a distance of 5m would raise the ball at constant velocity and now the ball would have 50J of gravitational potential energy.
Thinking in terms of the law of conservation of energy, the work done on the ball goes into an increase of gravitational potential energy. If I used a greater force and input more work, then the ball would gain kinetic energy in addition to the gravitational potential energy.