- #1
ownedbyphysics
- 16
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I need help with these problems that I'm not quite sure on how to do, I have attempted and gotten and answer, but I tend to be wrong...
1. a 0.095 KG pom pom slides along the gym flour with an initial speed of 5.50 m/s. It came to a complete stop in 1.9 m. What is the coefficient of friction between a pom pom and the gym floor?
a. I found acceleration by using a= vi^2/ 2d which is 5.50^2/ 1.9*2
b. then I used ma to find force then I assumed that F= the force of friction then isolated coefficient of friction and divided by mg using the equation Ff= umg and my answer was .811
I really didn't know what to do for the above problem, especially for step b, I just did whatever that came to mind, even though it doesn't really make much sense to me.
2. Michael stands on a scale in a moving elevator. His mass is 110.0 kg, the combined mass of the elevator and the scale is an additional 815.5 kg. Starting from rest, the elevator accelerates upward. During the acceleration, there is a tension of 9415.5 N in the hoisting cable. What is the reading on the scale (his weight) during the acceleration?
a. I found acceleration by using Ft= ma + mg = 9415.5N- (915.5)(9.81)= (915.5)a which was .4745 m/s^2
b. then I used F=ma so I mutlipled .475 by 110.0 kg and got 52.2 N
I have no idea if any of my answers make sense, please help!
1. a 0.095 KG pom pom slides along the gym flour with an initial speed of 5.50 m/s. It came to a complete stop in 1.9 m. What is the coefficient of friction between a pom pom and the gym floor?
a. I found acceleration by using a= vi^2/ 2d which is 5.50^2/ 1.9*2
b. then I used ma to find force then I assumed that F= the force of friction then isolated coefficient of friction and divided by mg using the equation Ff= umg and my answer was .811
I really didn't know what to do for the above problem, especially for step b, I just did whatever that came to mind, even though it doesn't really make much sense to me.
2. Michael stands on a scale in a moving elevator. His mass is 110.0 kg, the combined mass of the elevator and the scale is an additional 815.5 kg. Starting from rest, the elevator accelerates upward. During the acceleration, there is a tension of 9415.5 N in the hoisting cable. What is the reading on the scale (his weight) during the acceleration?
a. I found acceleration by using Ft= ma + mg = 9415.5N- (915.5)(9.81)= (915.5)a which was .4745 m/s^2
b. then I used F=ma so I mutlipled .475 by 110.0 kg and got 52.2 N
I have no idea if any of my answers make sense, please help!