I admit, I am a thorough novice in the field of physics but I have a question for those of you who are a little more learned in the subject. If a watch having a mass of 198 grams is dropped on to a sidewalk 35 ft. below, what force will it impact the sidewalk with? I thought that a simple force formula of F=mg [neglecting air resistance] would work for this, and using said formula I calculated the impact force to be approximately 2 newtons or 1.4 ft/lbs. Was I right in my calculation or are there other factors I have neglected to enter in?

No external forces , Energy is conserved :

0 + mgh = 0.5mV^2 + 0

Replace each value and find V .

I'm sorry, but could you a little more clear? I assume that mgh stands for mass(gravity)(height), but could you explain the formula you submitted in a little more detail?

First you might have to find the velocity just before hitting. And you should notice that you are told to find the force of impact. By assuming that the collision is inelastic and velocity after collision is zero, you can use Impact force= mass(velocity)/(time of impact)

But in the question, I can't see this figure given.

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Going to face the Hong Kong Advanced-Level Examination which is believed to be the most difficult AL exams in the world this year. I hope good results can be achieved.

Using just F = mg just gives you the weight of the watch if it were simply placed on the sidewalk, but that's not the force it impacts the sidewalk with. Also converting from Newtons to something in ft/lb is not correct.

Newtons are a unit of force, so the conversion should be just from Newtons to pounds. A ft-lb is energy, just like N-m. The final energy the watch has just as it strikes the sidewalk can be found, or put another way, the energy absorbed and converted to heat energy by the sidewalk can be found, but you lack the info to find the force the watch applies upon the sidewalk once it strikes the ground.

You need to know how quickly the watch changes it's velocity from it's maximum to zero. Here's why:

A change in momentum yields an impulse on the object that changed it's momentum. Given by:

mass*deltaV = Force*deltaT

Where deltaV is the change in velocity from striking the ground and deltaT is the time that it took to change that velocity from what it was just before it struck the ground to the final velocity of zero after it has struck the ground.

So the force it impacts the ground with is:

F = mass*(deltaV/deltaT)

You might think that deltaV/deltaT is an acceleration and you'd be correct. It's just not the acceleration due to gravity. It's the acceleration the watch undergoes from changing its velocity from its max velocity to zero velocity divided by the time that it took to make that velocity change.

Since it has a pretty fair velocity just before it strikes the ground which then goes to zero very quickly it undergoes a far greater acceleration than that of gravity upon impacting on the sidewalk. So the force it applies upon the sidewalk is initially far greater than the weight of the watch.

Without the time it took to change its velocity from some max value to zero, you can't solve for the force it impacts the ground with. You can find its max velocity, which it will have just before it strikes the ground, you can also find the energy the watch has as it impacts with the sidewalk, transferring that energy to the sidewalk which it converts to heat, but you lack the info to find the force of the impact. Sorry.

If you can find the time it took for the watch to change its velocity from its max just before striking the ground to zero just after impacting with the sidewalk then you can use the equation I gave you to find the force it strikes the sidewalk with, but you need that time.

You can't just assume it goes from some max velocity to zero in no time as that would yield an infinite impact force, so you need the non-zero time it took for the watch to change from a max velocity to zero velocity to find the impact force.

Remember that the weight some mass may have on Earth is not the same as the impact force that mass applies if dropped from some height.

Think of it like this, if the watch were simply placed on the top of your head you would feel the weight of the watch on your head, but it wouldn't be so great that it would hurt, but if that same watch is dropped from some high height and strikes you on the top of your head it's going to hurt like hell. That's because the impact force from being dropped from some high height upon your head is far larger than simply the weight of the watch.

Another way would be to find the rate of change in kinetic energy off the watch with respect to time just as the impact occurs. Here's why that would work:

KE = 0.5*mass*velocity^2 take the derivative of each side with respect to time:

dKE/dt = mass*dV/dt

That dV/dt is the same acceleration you need in the previous formula. That is dV/dt equals the deltaV/deltaT in the previous equation. In fact if you knew the value of dKE/dt that would be the impact force. The rate of change in kinetic energy with respect to time during the impact is directly equal to the product of the mass of the watch and the acceleration it undergoes from impacting the ground, that is the impact force.

So how quickly the mass transfers its kinetic energy into the sidewalk, with respect to the time it takes to transfer that energy, is also a measure of the impact force.

I know I didn't give you an actual answer to your question, but that's because I need that time to solve for that impact force.

Many Smiles,
Craig