mth174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Questions.
An athlete whose mass is 90.0kg is performing weight lifting exercises. Starting from rest position, he lifts, with a constant acceleration, a barbell that weighs 490N. He lifts the barbell a distance of .60m in 1.6s. Draw a clearly labeled free body diagrams for both. Use the diagrams and Newtons second law to find the total force that his feet exert on the ground as he lifts the barbell.
-So let's start by finding the acceleration that the barbell movies with: Taking average velocity, and knowing that its at rest and has a constant acceleration, then we can find .60m/1.6s= .375m/s,vf= .75m/s
a= .75m/s / 1.6s= .47 m/s^2
We can also go ahead and find the mass of the barbell, which 490N/9.81m/s^2, gives a 49.9kg barbell. Taking the mass and acceleration of the barbell, 49.9kg * .47m/s^2 we get 23.5N force that is moving the barbell upwards. But we want to find the force that the man applies to the ground in the course of his motion. (Here's where I want to check my reasoning) Would we take the mass of man, and multiply it by gravity to find the net force that is acting on the man, in this case its 90kg * 9.81m/s^2 which is 883N. So would would then take that and add it with the net force during the course of the man moving the barbell? 883N + 23.5N= 906.5 N
Gravity is also acting on the barbell. The net force on the barbell must be 23.5 N; however gravity is exerted a force of -490 N on it. In order to achieve the net force, what force must the weightlifter exert on the barbell? How does this affect your answers?
You have just landed on Planet X. You take out a 100-g ball and release it from rest at a height of 10.0m and measure that it takes 2.2s to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the 100 g ball weigh on the surface of Planet X.
Actually, once I rethought this question over it clicks a little bit more than it previously did before when I read it. So anyway, let's just check my reasoning skills to be sure that what I'm thinking is correct.
Since we're releasing the ball from rest, we can easily find the final velocity, which in this case is going to be the average velocity doubled: 10m/2.2s= 4.5m/s, so vf = 9.0m/s, then aave= 9m/s-0 / 2.2s = 4.09m/s^2. Then the acceleration due to gravity here is 4.09m/s^2, so the 100 g ball will simply be 100 * 4.09m/s^2
Very good, but 100 * 4.09m/s^2 = 409 m/s^2, which is an acceleration, not a weight.
What are the units of the 100? What therefore is the weight?
If we want to prove whether the integral of 1/t^ .3 from t=0 to t=1 converges or diverges, after integrating we can say that we want to find the limit as x->0 of 10/7 t^.7 from 1 to x. Then we just plug in and get 10/7 (1 ^ .7 ) - 10/7 (x ^.7 ). Since x will just go to zero, we get a numerical answer of 1.43, so it converges.
Yes. Very good.
Bounded by y= 4-x^2, y=0, x=-2, x=0, rotated about the x-axis. I know how to do these problems, but the most important part is finding the correct volume of a slice. This is an odd problem, and since I can correctly integrate this function, my error is probably lying in my volume of a slice. (Also, when we are given the x and y equal to does that mean that my integration bounds are going to be from -2 to 0 when I write my integral of this solid? Can I use pi* (4-x^2)^2 * d'x --would this be correct for the radius? But I know that we'd take the Riemann sum, then the integral, resulting (assuming this is correct) in integral( pi * (4-x^2)^2 dx from 0 to -2
Very good and correct throughout. The integral would be from -2 to 0, as you stated first, not from 0 to -2.
Find the volume of the solid whose base curve y= (x+e) -.5 between x=0 to x= .7 and y axis whose cross sections are perpendicular to the y axis are right triangles with altitudes 2.5 times base
This one is being rotated about the y-axis, right? For our volume slice we are using right triangles whose altitudes are 2.5 times the base, so would our height be 2.5h when we draw a facsimile of the triangle to represent it? Would the volume of a slice be s^2* d' h (thickness). Would we need to use similar triangles to come to a correct value for s? Would the width of the the function be to x= .7, and when considering a volume of a slice would we use the proportionality... I'm kind of stumped here
The cross-sections are perpendicular to the y-axis, so they run from the y axis to the curve, in the x direction.
Assuming the function is y = (x + e)^(-.5), to find a point on the curve which corresponds to the given y coordinate you would have to solve the given the equation for x; easy enough to do, and you get x = y^-2 - e. The line segment running from the y axis to this x coordinate therefore has length y^-2 - e.
Assuming this segment to be the base of the right triangle, its altitude would be 2.5 (y^-2 - e) and its area would be 1.2 * 2.5 (y^-2 - e) * ( y^-2 - e).
If the segment is the altitude of the right triangle, then the area of the cross-second would be 1.2 / 2.5 * (y^-2 - e) * ( y^-2 - e).
The Riemann sum is then set up, partitioning the y axis from y = e^(-.5) to (.7 + e)^-.5.