Question

A mass M of 3.60E-1 kg slides inside a hoop of radius R=1.30 m with negligible friction. When M is at the top, it has a speed of 5.17 m/s. Calculate size of the force with which the M pushes on the hoop when M is at an angle of 32.0 degrees.

Answer 1

Amyvincent are u a memorial student?

Answer 2

yes i am

Answer 3

lol basically we solving the last question loooooooooooool,i cant get through with this

Answer 4

physics 1050

Answer 5

lon capas are so confusing. I even looked up the methods for each question on google and yet my answers still come up incorrect. and yes, physics 1050.

Answer 6

i have a link i can give u,but after looking at it i can still really get it right

Answer 7

ok

Answer 8

http://www.google.ca/url?sa=t&rct=j&q=.%20calculate%20size%20of%20the%20force%20with%20which%20the%20m%20pushes%20on%20the%20hoop%20when%20m%20is%20at%20an%20angle%20of%2032.0%20degrees.&source=web&cd=2&ved=0CDUQFjAB&url=http%3A%2F%2Fwww.physicsforums.com%2Farchive%2Findex.php%2Ft-52200.html&ei=LlNST8SgB8Xg0QH5rfDDDQ&usg=AFQjCNHHa1o4XgsfXqqumpB-NL5RDT_aRg&cad=rja

Answer 9

i'll see if i can figure it out, what topics of the questions don't you understand? i might have some of them done.

Answer 10

i have all of them done except this last one

Answer 11

i have idea but am still not arriving at the answer,have 4 trials left

Answer 12

using this method
Etop = mv^2 + mgh
= 1/2)(.7)(5.75)^2 + .7(9.8)(2.8)
= 30.78 J
I
Etop = E43degrees
30.78 = .5(.7)(v^2) + .7(1.30(1-cos theta)(9.8)
v = 8.32m/s

Fnet = mv^2/R
mg - Fn = mv^2/R
Fn = mv^2/r - mgcos43
Fn = .7(8.32)^2/1.4 - .7(9.8)cos43
Fn= 29.59 N

Answer 13

using this method,am still getting 17.45N

Answer 14

i got my answer correct.. i'll tell you the formulas i used in order..

mv^2+mgh = (1/2)mv^2+mgh where h = 2R
where the first v^2 is the speed given and the right side of the equation speed is what you're trying to find. once you find that speed, fill it in to
FN= (mv^2)/R + mgcos\[\theta\]

Answer 15

sorry the theta should be on the line above but my computer is weird

Answer 16

what are your givens?

Answer 17

m=o.36
r=1.30
v=5.17
theta=32 deg

Answer 18

Am GETTING 12.84 N

Answer 19

oh! i have gotten it right
the problem was here :Fn = mv^2/r - mgcos43
it should be +

Answer 20

Sorry its 10.386

Answer 21

i got the answer as 23.44N,it accepted my answer

Answer 22

thanks vincent

Answer 23

perfect, would you mind helping me with another one of my questions i don't have done?

Answer 24

is it the S1 AND S2 pellet? LOL
this :When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.19 m/s. The mass stops a distance S2 = 1.85 m along the level part of the slide. The distance S1 = 1.17 m and the angle theta = 29.1 degrees. Calculate the coefficient of kinetic friction for the mass on the surface. ?

Answer 25

yes that one!

Answer 26

ok

Answer 27

what equations did you use?