By Norbury J.W.
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Additional info for Solutions manual for elementary mechanics and thermodynamics
Sample text
C) For what period t, will the mass m2 be at rest? D) If the masses are equal, what is the answer to Part C)? 81 m, what is the numerical value of this period? 48 CHAPTER 5. FORCE & MOTION - II SOLUTION Forces on m2 : Forces on m1 : Fy = m2 ay Fx = m1 ax T − W 2 = m2 a T = m1 v2 m1 (2πR/t)2 = R R m1 4π 2 R = t2 where we have chosen m2 a and T with the same sign. e. 81 m sec 49 7. A) What friction force is required to stop a block of mass m moving at speed v0 , assuming that we want the block to stop over a distance d?
A skier starts from rest at the top of a frictionless ski slope of height H and inclined at an angle θ to the horizontal. At the bottom of the slope the surface changes to horizontal and has a coefficient of kinetic friction µk between the horizontal surface and the skis. Derive a formula for the distance d that the skier travels on the horizontal surface before coming to a stop. (Assume that there is a constant deceleration on the horizontal surface). Check that your answer has the correct units.
We obtain t from the y direction 1 y − y0 = v0y t + ay t2 2 Now y0 = 0, y = −H, v0y = 0, ay = −g giving 1 −H = − gt2 2 or t= 2H g Substuting we get R = v0 t = v0 2H g Check units: The units of v0 2H g are √ m −1 = m sec sec2 = m sec−1 sec = m m sec−2 which are the correct units for distance. m sec−1 31 10. A skier starts from rest at the top of a frictionless ski slope of height H and inclined at an angle θ to the horizontal. At the bottom of the slope the surface changes to horizontal and has a coefficient of kinetic friction µk between the horizontal surface and the skis.
Solutions manual for elementary mechanics and thermodynamics by Norbury J.W.
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