# 2 Kinetic Questions

Peach

## Homework Statement

While doing a chin-up, a man lifts his body 0.40 m. The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.)

W = F * s

## The Attempt at a Solution

m = w/(g*s)
m = 70J/(9.8m/s^2 * .40m
m = 17.86 kg

Then I take this and divided by total mass 70kg to get the percentage but I got the wrong answer. I don't know why...what am I doing wrong here?

## Homework Statement

Rotating Bar. A thin, uniform 12.0-kg bar that is 2.00 m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar?

## Homework Equations

KE = 1/2m (vf^2 - vo^2)

## The Attempt at a Solution

I didn't know how to get this started but I gave it a shot. First, I took the 5 revolution per 3 seconds and converted them to velocity.

5rev/3s (2*pie*r/rev) = 20.9 m/s

Then I take that answer and plug it into the kinetic equation but as expected, I was wrong. This kinetic stuff is really confusing or is it just me? Pls help?

Last edited:

Homework Helper

## Homework Statement

Rotating Bar. A thin, uniform 12.0-kg bar that is 2.00 m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds.

What is the problem question here?

Peach
Oops. I totally forgot.

Question: What is the kinetic energy of this bar?

l46kok

## Homework Statement

While doing a chin-up, a man lifts his body 0.40 m. The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.)

W = F * s

## The Attempt at a Solution

m = w/(g*s)
m = 70J/(9.8m/s^2 * .40m
m = 17.86 kg

Check your math. Your equation seems right, but I get a different answer when I plug in the values.

Homework Helper
Oops. I totally forgot.

Question: What is the kinetic energy of this bar?

Excellent. Now, how can you calculate the angular velocity of the bar from the given information? The units should be [rad/s].

Further on, what is the definition of rotational kinetic energy?

l46kok

## Homework Statement

Rotating Bar. A thin, uniform 12.0-kg bar that is 2.00 m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar?

## Homework Equations

KE = 1/2m (vf^2 - vo^2)

## The Attempt at a Solution

I didn't know how to get this started but I gave it a shot. First, I took the 5 revolution per 3 seconds and converted them to velocity.

5rev/3s (2*pie*r/rev) = 20.9 m/s

Then I take that answer and plug it into the kinetic equation but as expected, I was wrong. This kinetic stuff is really confusing or is it just me? Pls help?

First of all, your radius of the bar is 1m. Don't get confused.

Second of all, look into Rotational Kinetic Energy. You're using the linear kinetic energy equation

Last edited:
Peach
1) I checked my math already and I still get 17.86kg. It should be 70/3.92 right? Are you sure I'm not missing anything?

2) I haven't learned about rotational kinetic energy yet. That's on chapter 9, I'm only on chapter 6.

Edit: Is the angular speed (5rev/3s)(2pie/rev) = (10pie*rad/3s)?

Last edited:
Peach
Can someone pls pls pls help me with the first problem? I can't figure it out at all, I don't know where I went wrong. I've been punching numbers for the last hour and I still can't seem to get the right answer and this, I know, is a really easy problem and yet I'm going crazy over it. Sigh.

1. Homework Statement
While doing a chin-up, a man lifts his body 0.40 m. The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.)

2. Homework Equations
W = F * s

3. The Attempt at a Solution
m = w/(g*s)
m = 70J/(9.8m/s^2 * .40m
m = 17.86 kg

Mentor
First problem

## Homework Statement

While doing a chin-up, a man lifts his body 0.40 m. The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.)

W = F * s

## The Attempt at a Solution

m = w/(g*s)
m = 70J/(9.8m/s^2 * .40m
m = 17.86 kg

Then I take this and divided by total mass 70kg to get the percentage but I got the wrong answer. I don't know why...what am I doing wrong here?
Not quite sure what you are doing here. Try this. Call the man's total mass M and his muscle mass m. Figure out: The amount of work required to do one chin up; the amount of work his muscle mass can generate. Set them equal and solve for m/M.