# Help with problems over Newton's Laws

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

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

PhanthomJay
Homework Helper
Gold Member
ownedbyphysics said:
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

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

Good start on (1), you found the acceleration correctly, and corectly determined the friction coefficient. But what you MUST do is draw free body diagrams. Examine all the forces acting. Only friction in this case, so that's the total net force causing the acceleration.

In (2), you again have solved a correctly , now draw a free body of Michael...his weight acting down, the normal force (scale reading) acting up, and a you have solved already up. Use newton 2 again to solve for the normal force.

OlderDan
Homework Helper
I think your #1 is OK. You are on the right track with #2, but you have an addition problem and maybe some typos in your post. The net force on Michael is not what he will read on the scale.

for the addition problem, it was the formula that my teacher told us for Fty...so I used that for the elevator problem since it's up and down. I tried 52.2N= m(9.81) = 5.32kg, but that doesn't make any sense. I'm sorry but I have no idea what to do next!

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or maybe I have to find the net acceleration since I found the acceleration acting up so would I do 9.81- .4745 =9.3355 then multiply it by the mass to get the weight? = 1026.905 N? would the scale be reading newtons?

OlderDan
Homework Helper
ownedbyphysics said:
for the addition problem, it was the formula that my teacher told us for Fty...so I used that for the elevator problem since it's up and down. I tried 52.2N= m(9.81) = 5.32kg, but that doesn't make any sense. I'm sorry but I have no idea what to do next!
The addition problem is adding the masses. The total mass is 925.5kg, not 915.5kg. That changes the weight of the elevator and contents, which changes the difference between tension and weight, which changes the acceleration. I think your method is OK, but when you find the final net force on Michael you have to remember that the scale is not reading the net force; it is reading the normal force.

OlderDan said:
The addition problem is adding the masses. The total mass is 925.5kg, not 915.5kg. That changes the weight of the elevator and contents, which changes the difference between tension and weight, which changes the acceleration. I think your method is OK, but when you find the final net force on Michael you have to remember that the scale is not reading the net force; it is reading the normal force.

That's the part I don't understand because Fn= mg....I don't know how to find the normal force without just using 110.0*9.81

OlderDan