I have a question about Net force and Gravity. For Example, If you were to throw a ball (1 kg) straight up in the air at a starting velocity of 10 m/s, without air resistance, what’s the net fore that acts on the ball half way up its path?
Now I looked at this a few different ways but I'm still puzzled.
If the starting velocity is 10 m/s (without air resistance) and the pull of gravity is -9.8m/s then the net force would be .2 m/s? That didn’t seem nearly close to what it would be and know its wrong…I thought that would be by how much the balls velocity decreases by each second.
Then I realized that the ball is going in a arch. At the top of its path it will be zero. The gravity and the balls velocity will be completely balanced. But half way up the path the forces will not be balanced…the balls velocity will over come gravity and in result be slowed down. Therefore, at the middle of the path the net force would be:
Ball: 5m/s >< 9.8 G which is constant. Then it would be 4.8m/s? But...that doesn’t sound right… If anyone could help me with is problem I’m having that would be most appreciative!!!!
You're muddling up a lot of stuff there.
g is an acceleration, with units of m/s², while velocity has units of m/s. You therefore can't consider their difference, because it's a meaningless expression.
The force on an object of mass m is (at least over small distances close to the Earth's surface) mg.
Using Newton's 2nd law you have F=ma and F = -mg, therefore a = -g. The minus sign is because I'm measuring distance up and gravity pulls you down.
You therefore have a = -g
The change in an objects velocity in t sections is therefore just at = -gt.
If a ball starts at 10m/s upwards, after t seconds it'll be at 10+(-gt) = 10-gt m/s.
Can you see how I've related the 10m/s and the g = 9.8m/s² ? Since g is an acceleration, it needs to be multiplied by some multiple of t to get the right units.
The time the ball stops is when 10-gt = 0, so t = 10/g = 1.02 seconds.
The acceleration is ALWAYS -g because the force is a constant -mg. It is not the acceleration which changes, but the ball's velocity.
Does that help? Feel free to ask for any clarification or elaboration on any part of that
g is an acceleration, with units of m/s², while velocity has units of m/s. You therefore can't consider their difference, because it's a meaningless expression.
The force on an object of mass m is (at least over small distances close to the Earth's surface) mg.
Using Newton's 2nd law you have F=ma and F = -mg, therefore a = -g. The minus sign is because I'm measuring distance up and gravity pulls you down.
You therefore have a = -g
The change in an objects velocity in t sections is therefore just at = -gt.
If a ball starts at 10m/s upwards, after t seconds it'll be at 10+(-gt) = 10-gt m/s.
Can you see how I've related the 10m/s and the g = 9.8m/s² ? Since g is an acceleration, it needs to be multiplied by some multiple of t to get the right units.
The time the ball stops is when 10-gt = 0, so t = 10/g = 1.02 seconds.
The acceleration is ALWAYS -g because the force is a constant -mg. It is not the acceleration which changes, but the ball's velocity.
Does that help? Feel free to ask for any clarification or elaboration on any part of that
The force is constant at 9.81 newtons, at the top, at the bottom, half way up, the force is constant. Other things change: time, distance, velocity, direction, but force is constant. F = ma; 9.81 newtons = 1 kg * 9.81m/sec/sec, this means acceleration is constant as well.
The fact that it is going up might be what is confusing you, d = ½ at² or since t =v/a d = ½ v²/a = 5.097 m = ½ *10*10/9.81 it rises 5.097 m it takes 1.02 sec. Start by dropping it from 5.097 m, maybe that will make scene.
The fact that it is going up might be what is confusing you, d = ½ at² or since t =v/a d = ½ v²/a = 5.097 m = ½ *10*10/9.81 it rises 5.097 m it takes 1.02 sec. Start by dropping it from 5.097 m, maybe that will make scene.