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blaziken's_charizard
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1. Explain why the weight of a body changes if it is taken form the equator to one of the poles
2. With Reference to Newton’s Laws of Motion, determine the force that a man of mass 75.0kg exerts on the bottom of an elevator when it is
a. Ascending with a constant velocity of 2.5m/s
b. Descending with an acceleration of 4.0m/s2
3. A 100kg pile driver is suspended 12.0m above the ground where it is eventually released and falls, driving a pile into the ground
a. What is the gravitational potential energy before it is released and ½ way of its fall
b. Why the difference in gravitational potential energy above
[bThe attempt at a solution:[/b]
1. I will have to research question one, but any help would be appreciated.
2. Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. Newton's second law can be summarized by the formula:
f=ma
a) When ascending: f=75kg*2.5m/s^2 = 187.5 N
b) When descending: f=75kg*4m/s^2 = 300 N
I have a feeling that I am wrong because I didn't consider the motion of the elevator...
G.P.E=mgh= 100kg*10m/s^2*12m = 12000 J
G.P.E at half the distance= 1/2*12000J= 6000 J
There is a difference of the two G.P.E's above because G.P.E is defined by the formula: mass(m)*gravitational acceleration(g)*height(h). At half-way during the fall, the distance is also half so as a result the value of mgh would now be mgh/2 as opposed to the GPE just before the fall (mgh), and hence the difference in the two.
I think I have done a fair effort and it would be even better if I am corrected and guided, thank you.
2. With Reference to Newton’s Laws of Motion, determine the force that a man of mass 75.0kg exerts on the bottom of an elevator when it is
a. Ascending with a constant velocity of 2.5m/s
b. Descending with an acceleration of 4.0m/s2
3. A 100kg pile driver is suspended 12.0m above the ground where it is eventually released and falls, driving a pile into the ground
a. What is the gravitational potential energy before it is released and ½ way of its fall
b. Why the difference in gravitational potential energy above
[bThe attempt at a solution:[/b]
1. I will have to research question one, but any help would be appreciated.
2. Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. Newton's second law can be summarized by the formula:
f=ma
a) When ascending: f=75kg*2.5m/s^2 = 187.5 N
b) When descending: f=75kg*4m/s^2 = 300 N
I have a feeling that I am wrong because I didn't consider the motion of the elevator...
G.P.E=mgh= 100kg*10m/s^2*12m = 12000 J
G.P.E at half the distance= 1/2*12000J= 6000 J
There is a difference of the two G.P.E's above because G.P.E is defined by the formula: mass(m)*gravitational acceleration(g)*height(h). At half-way during the fall, the distance is also half so as a result the value of mgh would now be mgh/2 as opposed to the GPE just before the fall (mgh), and hence the difference in the two.
I think I have done a fair effort and it would be even better if I am corrected and guided, thank you.