Crosswords1 min ago
How do I calculate this force?
If I drop an object (particle), that has a mass of 5kg, from a height of 1.5m, onto someone's foot, how do I calculate the force that it exerts onto the foot?
Correct answers only please.
Correct answers only please.
Answers
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Force is rate of change of momentum. The speed of the object the instant before hitting the foot will be 5.4m/s (using the SUVAT equations) It's momentum will therefore be 27Ns. (p=mv)
To calculate the force on the foot you need to know the time taken for the mass to come to rest. If this was, say, 0.01s the force on the foot would be 2700N. (F = change in p divided by change in time)
To calculate the force on the foot you need to know the time taken for the mass to come to rest. If this was, say, 0.01s the force on the foot would be 2700N. (F = change in p divided by change in time)
The weight of a 5kg object in Earth's gravitational field is 5kg X 9.81 m/s squared = 49N.
The force of the object when it hits the foot is entirely different. The object will be travelling at 5.4m/s immediately before it hits the foot. The object will then slow down (decelerate) and come to a stop. A force will be applied to the foot which is the rate of change of momentum. The greater the height, the greater the rate of change of momentum and, consequently, the greater the force on the foot.
The force of the object when it hits the foot is entirely different. The object will be travelling at 5.4m/s immediately before it hits the foot. The object will then slow down (decelerate) and come to a stop. A force will be applied to the foot which is the rate of change of momentum. The greater the height, the greater the rate of change of momentum and, consequently, the greater the force on the foot.
A falling object will have kinetic energy, which will be imparted to your foot on impact � once the impact energy has been absorbed and the object come to rest, it will apply a force to your foot.
Factor30�s first answer gives this force in Newtons.
The energy is given by the formula E = H x g x m where:-
E= energy in Joules
H = height through which the object falls in metres
g = gravitational acceleration in m/s/s (9.81)
m = mass of object in kilograms.
Therefore we get E = 1.5 x 9.81 x 5 = 73.6 Joules
Energy and Force are two different things.
Factor30�s first answer gives this force in Newtons.
The energy is given by the formula E = H x g x m where:-
E= energy in Joules
H = height through which the object falls in metres
g = gravitational acceleration in m/s/s (9.81)
m = mass of object in kilograms.
Therefore we get E = 1.5 x 9.81 x 5 = 73.6 Joules
Energy and Force are two different things.
Thanks Hymie.
Yes, the force is 49 Newtons. This is the force whatever height from which the 1.5 kg weight is dropped.
And yes, the kinetic energy that increases with speed, and for a falling object the speed on impact depends on the height from which it is dropped. Teddio's figure of 5.4 m/s after falling 1.5 metres is correct
So Shaun, if this is a homework question which asks for the force on impact the answer given by me and Hymie of 49N is correct. If the question specifically asks about energy then Teddio's answer becomes relevant use the energy formula given by Hymie or use the kinetic energy formula KE=0.5 x mass x v^2 which gives the same result.
Yes, the force is 49 Newtons. This is the force whatever height from which the 1.5 kg weight is dropped.
And yes, the kinetic energy that increases with speed, and for a falling object the speed on impact depends on the height from which it is dropped. Teddio's figure of 5.4 m/s after falling 1.5 metres is correct
So Shaun, if this is a homework question which asks for the force on impact the answer given by me and Hymie of 49N is correct. If the question specifically asks about energy then Teddio's answer becomes relevant use the energy formula given by Hymie or use the kinetic energy formula KE=0.5 x mass x v^2 which gives the same result.
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I don't know what you mean Shaun when you say you made a stupid mistake. Your 'stupid mistake' method was the way I did it which gave an answer of 49N. I think that method is in fact correct, and no-one has quoted a different value for the force
I think you are saying that rather than consider the force (F=mg) immediately before it hits your foot you should use the equation F= ma [same as F= m (v-u) /t] based on the deceleration after it hits your foot. Are you saying that you eventually came up with a different answer- and if so, what was it?
Instinctively I still feel the Force acting on your foot is the 49N.
Cheers
I think you are saying that rather than consider the force (F=mg) immediately before it hits your foot you should use the equation F= ma [same as F= m (v-u) /t] based on the deceleration after it hits your foot. Are you saying that you eventually came up with a different answer- and if so, what was it?
Instinctively I still feel the Force acting on your foot is the 49N.
Cheers
Factor 30: there is insufficient data in Shaun's question to calculate the force on the foot. You need a value for the deceleration or the stopping time. If the mass was placed on the foot, ie. it was stationary, the downwards force would be the weight (49N). Because the mass is moving when it hits the foot, the force must be greater than 49N. My original calculation assumed a stopping time of one hundreth of a second. For a shorter stopping time the force would be greater than 2700N.
Thanks Teddio. Do we agree that the downwards force associated with the falling weight is 47N immediately prior to impact. This would be the same whatever height it was dropped from.
I can see the height will affect the speed at the point of impact but I can't see an equation which would show how the speed affects the force, since F=mg.
I'll have to think about your other point which is that the force during impact depends on the time taken for the weight to slow down after impact- the cushioning effect. I admit I'm no physicist but I'll look it up
I can see the height will affect the speed at the point of impact but I can't see an equation which would show how the speed affects the force, since F=mg.
I'll have to think about your other point which is that the force during impact depends on the time taken for the weight to slow down after impact- the cushioning effect. I admit I'm no physicist but I'll look it up
Thanks Shaun.
Yes the brick dropped from a height will have greater effect than a brick placed on my head.
That's because the brick that is dropped from a building will have a higher momentum (mass x velocity) than the stationery brick (momentum =0). It will also have higher kinetic energy (0.5 x m x v^2).
But momentum and KE are not the same thing as force. While the brick is falling I maintain that the force acting downwards is always 49 Newtons, irrespective of the height from which it was dropped. Is there a formula which disproves this?
Yes the brick dropped from a height will have greater effect than a brick placed on my head.
That's because the brick that is dropped from a building will have a higher momentum (mass x velocity) than the stationery brick (momentum =0). It will also have higher kinetic energy (0.5 x m x v^2).
But momentum and KE are not the same thing as force. While the brick is falling I maintain that the force acting downwards is always 49 Newtons, irrespective of the height from which it was dropped. Is there a formula which disproves this?
factor30 wrote: "While the brick is falling I maintain that the force acting downwards is always 49 Newtons, irrespective of the height from which it was dropped."
I agree with that, BUT... oh nevermind, I've confused myself now. factor30 said "I admit I'm no physicist ". That's the embarrasing thing you see - I am believe it or not. I've had some time out from it and I seem to have forgotten a lot. It's like they say - Use it or Lose it.
I agree with that, BUT... oh nevermind, I've confused myself now. factor30 said "I admit I'm no physicist ". That's the embarrasing thing you see - I am believe it or not. I've had some time out from it and I seem to have forgotten a lot. It's like they say - Use it or Lose it.
Teddio wrote "Because the mass is moving when it hits the foot, the force must be greater than 49N. My original calculation assumed a stopping time of one hundreth of a second. For a shorter stopping time the force would be greater than 2700N"
I agree. Although I have agreed in the previous reply that the only downwards force is the weight of the particle, 49N.
So where does this extra force come from??
I agree. Although I have agreed in the previous reply that the only downwards force is the weight of the particle, 49N.
So where does this extra force come from??
I can't help thinking that most people here are missing the point.
It is true that while the object is falling under gravity there is a net downward force (owing to gravity) of 49N (ignoring air resistance for now), but as soon as it touches the foot then the foot exerts an upward force on the object sufficient to stop it in a relatively small distance.
We seem agreed that the object is moving at 5.4m/s when it strikes the foot. Assuming that it stops in a distance of about 1cm then by using v^2=u^2+2fs we can calculate that there is a net upward acceleration of 1458m/s^2 causing the object to stop. Adding to this the acceleration due to gravity (gravity is still trying to accelerate it) we get a gross upward acceleration of 1467.8m/s^2.
This means that the foot exerts an upward force on the object of 7339N. Or it could be said (incorrectly) that the weight exerts that same force downwards on the foot.
To put this in perspective, and in a way that most people will comprehend, this is equivalent to nearly 150 times the force of gravity.
It is true that while the object is falling under gravity there is a net downward force (owing to gravity) of 49N (ignoring air resistance for now), but as soon as it touches the foot then the foot exerts an upward force on the object sufficient to stop it in a relatively small distance.
We seem agreed that the object is moving at 5.4m/s when it strikes the foot. Assuming that it stops in a distance of about 1cm then by using v^2=u^2+2fs we can calculate that there is a net upward acceleration of 1458m/s^2 causing the object to stop. Adding to this the acceleration due to gravity (gravity is still trying to accelerate it) we get a gross upward acceleration of 1467.8m/s^2.
This means that the foot exerts an upward force on the object of 7339N. Or it could be said (incorrectly) that the weight exerts that same force downwards on the foot.
To put this in perspective, and in a way that most people will comprehend, this is equivalent to nearly 150 times the force of gravity.
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