Question

3. A mass mis initially at rest at a height h above the "ground" along an incline as seen in the diagram. The mass is then released and slides down the incline, and then along a horizontal surface before it is seen to rise to a height of only one half the initial height along the opposite incline. Find an expression for the coefficient of friction in terms of the angle o only, and show it can be written as 3 coto +2 Keep in mind that friction acts along all 3 surfaces - both inelines and the flat surface in between Solve this problem using a combination of forces and kinematics, and be sure to explain your steps. To make this problem easier, you should should treat this as 3 separate problems by breaking it down into 3 different regions: 1) the way down resulting in a velocity due to an acceleration of ai), 2) across the horizontal ending with a velocity 1 (due to an acceleration as), 3) on the way up stopping at a height 1/2 (due to an acceleration as). Keep in mind that the final velocity of one region becomes the initial velocity of the next region. It is also much better to deal with the squares of the velocity rather than the velocity itself. Figure 3: Inot na Duh F=ma MEN 0-n. You should find the following: .Az a = √ a g(sino - 03) = -9(sin $ + picos) - 9 sino – 4,0 4), *- --9--- in o + 4.056) – 2011 – micot 6), @-29h(1 – ---cote)

Answer 1

ngtad mg Sint h/2 D=hmg Cohen mau caw at husht h- then mg Sinp-Hmgcerma a=g(Sint-Mcoi ) be Speed al he bottom Led V=2ax b/Smt 2 gLSmb-co) /sint Vi2gh-H colt) Cohen ma wai at boltom Ld Vbe speed wt bejere gumg up. V22-v2= 2 a, h 2gh(1- Lott)- 2ugh 2=2gh -HCott- V22 Cohen ha Cwa o1ihs A2g Sn-gcoip 2 a2/2/m 2gh(1-H- Lolt) =gh (Smt +M) 2 (1-M-Mo)+Hcol 2-24-2HCotp= I+ Hcot T : 3Hcotb+ 2H 0V2 Sind 3colt +2