Motor Force Calculation for Remote Control Toy Car

[pivoxa15]

Homework Statement

A remote control electric toy car of mass 0.5kg starts from rest and reaches a speed of 1.8m/s in 2s while traveling south along a straight level path. During this time the average force needed to overcome friction is 0.25N.

What force is exerted by the motor of the car during this time? (Assume that this force is constant.)

Homework Equations

a=del(v)/del(t)

The Attempt at a Solution

The observed acceleration is .9m/s^2 hence an observed force of .45N forwards. The car in traveling must also provide a .25N in the same direction in order to eliminate friction so that comes to a total of .7N exerted by the motor. Correct?

The answer at the back of the book suggested .45N only. Is the answer wrong?

 

[pivoxa15]

While on this topic there is another similar question.

An object reaches a rough surface where the force of friction is 5N. The applied force on the object is 10N and the object is dragged over a distance of 9m.
What is the work done by the applied force?

Is it only 45N as only the force along the direction of movement of the object is counted? So the force gone into friction is not counted as work as that applied 5N did not go into moving the mass any distance.

 

[Dick]

You are correct.

 

[Dick]

I meant you are correct on the first question. On the second one, 45J (not N!) is correct but the reason you gave is wrong. The frictional force is exactly what counts. If 10N is exerted then presumably it is partially in a normal direction leaving only 5N as the horizontal component. Force only does work in the direction of motion and frictional forces are always in the direction of motion.

 

[pivoxa15]

I think the 10N was purely in the horizontal direction which is in the direction of motion.

I am sure that frictional force is always in the opposite direction of motion which is contray to what you say.

 

[Dick]

OK, opposite direction of motion. I expressed myself badly. Having everything horizontal can't work. Draw a diagram for the horizontal forces. 10N (push) in the direction of motion. 5N (friction) opposite. That's it. Those are ALL of the horizontal forces. I would then conclude the object must be accelerating and the extra 45J are going into KE - which doesn't really sound like 'dragging'...

 

[pivoxa15]

Actually I think the work done by the applied force is 90N, not 45J because W=Fd. The force applied is 10N even though 5N is gone into overcoming the friction.

I don't think the energy used in overcoming friction has gone into KE of the object. The system includes the ground and object. 45J into KE and 45J gone into heat in the ground. Total work applied is 90N.

 

[Hootenanny]

You are correct. The work done by the applied force is 90J, irrespective of any resistive forces. As you correctly say, 45J of work was done overcoming friction and 45J of work was done accelerating the object.

 

[Dick]

I agree completely. If you think the problem implies that the applied force is horizontal. I think the intention of the problem is that the 10N force be applied at an angle (with 5N horizontal component). That makes their given answer of 45J correct and is more consistent with my picture of 'dragging'. But in either case I think we understand the physics...