Recent content by Clockclocle

yes the answer is 2 (sqrt(2)+1). So is it true?

Here is my attempt. At ymax the velocity turn to zero so we get time t*=v0/g and ymax=1/2 (v0^2/g). At the height y max, since the velocity at this point is 0, i get another equation y= 1/2(v0^2/g)-(g/2)t^2, this equation could be considered as continuation of first equation. Set ymax/2=1/4...

I see the mistake, I thought that it gonna fall below me so it would be the same when I stand still. But in this case I keep moving with velocity v0.

In this situation should my free fall equation contain the v0 of the baloon or I should deny it. Because it seems to me that there is no outer force acts on the sandbag, so the scenario is just the same as I climb to the same height at time t=0 and drop the sandbag at rest.

The instantaneous velocity at time a is defined as derivative of motion function f(t). It is not similar to average velocity in an interval of time. From the Newton law. If an object is at rest, we must exert a force to make it move, assume that there is no friction. Depend on the weight of...

What? I think the last book would both contact with the surface and the top of it?

Suppose n book stack on each other. Since each book have the same weigh then the last book exert a force N=nmg on the surface so it has the biggest static friction. But if we treat the whole tower of books as one particle it also has N=nmg. This mean if we exert enough force in the last book...

When I am riding a motocycle, suppose I slowly accelerate the car(push the pedal) the net force >0 for an interval of time, when it gets maximum the car move with constant velocity and net force become zero. Does it mean friction "come after" the exert force on the car?

Does the kinetic friction monotonically deacrease to 0 as the velocity become 0? Does it make velocity become zero and disappear immediately it mean the graph of kinetic friction respect to time t after I realease the object is continuous?

Because It make nonsense if friction remain still, it has to be reduced to make sure the object don't move backward. Friction proportional to exert force. When I realease the object it still move a little bit before stop then I guess the friction have to be reduced to its static form as the...

Im looking for an easier explanation but it seem like 3 Newton laws don't provide enough about the friction. According to Lnewqban, The friction only reduce to 0 but not make it move backward, why is it? Is there a change of kinetic friction reduce to static friction as the velocity become zero

it mean the reaction force the surface exert back to object isn't a friction?

It mean when I stop pushing only the exert force disappear immediately but the friction remain still? I thought that kinetic friction is the reaction of the force which the object exert on the surface due to third law, why it don't disappear?

When I exert enough force that overcome the static friction. The object start moving and surface create kinetic friction on object if I exert harder overcome the maximum of friction it start accelerate. When I release the object will the kinetic friction disappear immediately proportional to...

Since the friction is applied from surface to object while object moving,it has its maximum depend on what kind of surface?