Newton's Laws - I need help


by sinisterlink
Tags: laws, newton
sinisterlink
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#1
Oct1-04, 07:16 PM
P: 9
An athlete whose mass is 93.0 is performing weight-lifting exercises. Starting from the rest position, he lifts, with constant acceleration, a barbell that weighs 500 . He lifts the barbell a distance of 0.700 in a time of 2.10 .

Use Newton's laws to find the total force that his feet exert on the ground as he lifts the barbell. Take free fall acceleration to be 9.80 .


The answer is 1430 N


Please show me how to get the answer. I have a physics test on Monday and I need help bad.
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Sirus
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#2
Oct1-04, 07:35 PM
P: 582
(Hopefully) no one will "show you how to get the answer"; rather, we are here to help you understand the concepts related to the question and help you figure it out so you can do it on your own. Please include the units of all the data in the question, and show us what you have done so far and where you are having problems.
sinisterlink
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#3
Oct1-04, 07:38 PM
P: 9
sorry, forgot to put in the units. My friend helped me out. But I'm completely lost. I dont know what to do first, this semester has been really tough on me.

An athlete whose mass is 93.0kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with constant acceleration, a barbell that weighs 500N . He lifts the barbell a distance of 0.700m in a time of 2.10s .

Use Newton's laws to find the total force that his feet exert on the ground as he lifts the barbell. Take free fall acceleration to be 9.80m/s^2 .


The answer is 1430 N

Tide
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#4
Oct1-04, 07:39 PM
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Newton's Laws - I need help


First, it is important to include units with each quantity!

You need to calculate the acceleration on the barbell which will tell you the force that was applied to it. That is the same force the barbell exerts on the weight lifter. That force plus his weight determine how much force his feet exert on the ground.
sinisterlink
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#5
Oct1-04, 07:47 PM
P: 9
How would I go about calculating the acceleration of the barbell?

Do I have to take gravity into account if they give me .7m and 2.1s?
Tide
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#6
Oct1-04, 08:03 PM
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Since you're given the acceleration is uniform, the height to which the weight is lifted and the time it takes to lift it then you only need:

[tex]h = \frac{1}{2} a t^2[/tex]

to find the acceleration.
sinisterlink
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#7
Oct1-04, 08:03 PM
P: 9
I have no clue how to get the accleration of the barbell. Everytime I try my final answer doesnt come out to 1430N.

i was using this formula: X(final) = X(start) + V(start)t + 1/2 at^2

.7 = 0 + 0(2.1) + 1/2a(2.1)^2
a= .3175

then F=ma

500(.3175)=158.73 + 93(9.8) = 1070.13

which is wrong. Help
tony873004
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#8
Oct1-04, 08:13 PM
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Are you sure about the 1430N as the right answer?

F=ma

F = 500 kg * 9.81 m/s^2

F = 4905 N for the barbell alone without the added acceleration of lifting it.

Or am I doing this wrong? These are the kinds of problems we're going to do in 2 weeks from now, so I have an interest in attempting this problem with you.
sinisterlink
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#9
Oct1-04, 08:16 PM
P: 9
is 1430 wrong ?

is it actually the answer I got above?

This is killing me, I have 9 more problems to go and I can't move on if I don't get this one right. I'm stubborn.
Pyrrhus
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#10
Oct1-04, 08:18 PM
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What's a barbell? i imagine is some kind of ball.
sinisterlink
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#11
Oct1-04, 08:20 PM
P: 9
A barbell is a pole that you do Bench Presses with.

He's lifting it off the ground, I imagine.
tony873004
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#12
Oct1-04, 08:21 PM
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I'm not sure if this is right, but I'd add the 0.317460317460317 that you got for acceleration to gravity's 9.81 for total acceleration of the barbell, figure out the force from that mass and acceration and add it to the force computed by f=ma for the weightlifter (93 * 9.81)
But this answer exceeds 5000, well above your 1430 expectation.
tony873004
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#13
Oct1-04, 08:23 PM
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Quote Quote by Cyclovenom
What's a barbell? i imagine is some kind of ball.
Imagine 2 bowling balls joined by a metal pole that's 6 feet long.
tony873004
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#14
Oct1-04, 08:24 PM
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How do you know the answer is 1430? Is that given in the back of the book? Or is this an online instant graded homework?
sinisterlink
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#15
Oct1-04, 08:29 PM
P: 9
I think I got it.


First I get the mass of the barbell by dividing 500 by 9.8

51.02

then i get the normal force the lifter uses.

Fnormal = m(a+g)

51.02(.3175+9.8)

Fnormal=516.195 + 93(9.8) = 1427.59

which is pretty close, if i didnt round i probbaly get 1430
tony873004
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#16
Oct1-04, 08:34 PM
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Quote Quote by sinisterlink
I think I got it.


First I get the mass of the barbell by dividing 500 by 9.8

51.02
But 500 is the mass of the barbell....??? Wait. You didn't put units in the original problem. It says the WEIGHT of the barbell is 500. But weight is not mass. And 500 kg is awfully heavy for a barbell. Can you restate the original question with units?
Pyrrhus
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#17
Oct1-04, 08:35 PM
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First of 500 N is not MASS, It's WEIGHT, now try again.
tony873004
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#18
Oct1-04, 08:48 PM
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I get the same answer as you, 1429, using 500 N instead of 500 kg.


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