226_PartUniversity Physics Solution

226_PartUniversity Physics Solution - Potential Energy and...

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Potential Energy and Energy Conservation 7-15 SET UP: Let point 1 be where it enters the rough bottom and point 2 be where it stops. 1 1 other 2 2 K UW KU + += + EXECUTE: other k , f WW m g s μ == 212 0; KUU = 1 78.4 J K = k 78.4 J 0; mgs −= solving for s gives 20.0 m. s = The wood stops after traveling 20.0 m along the rough bottom. (b) Friction does 78.4 J of work. EVALUATE: The piece of wood stops before it makes one trip across the rough bottom. The final mechanical energy is zero. The negative friction work takes away all the mechanical energy initially in the system. 7.48. IDENTIFY: Apply Eq.(7.14) to the rock. k other f = . SET UP: Let 0 y = at the foot of the hill, so 1 0 U = and 2 Um g h = , where h is the vertical height of the rock above the foot of the hill when it stops. EXECUTE: (a) At the maximum height, 2 0 K = . Eq.(7.14) gives k Bottom Top f K WU . 2 0k 1 cos 2 mv mg θ dm g h . sin dh θ = , so 2 1 cos 2s i n h vg g h μθ . 22 2 1c o s 4 0 (15 m/s) (0.20)(9.8 m/s ) (9.8 m/s ) i n 4 0 hh ° ° and 9.3 m h = . (b) Compare maximum static friction force to the weight component down the plane. 2 ss cos (0.75)(28 kg)(9.8 m/s )cos40 158 N fm g ° = . 2 s sin (28 kg)(9.8 m/s )(sin40 ) 176 N mg f = °= > , so the rock will slide down. (c) Use same procedure as in part (a), with 9.3 m h = and B v being the speed at the bottom of the hill. k Top B f K . 2 kB 1 cos sin 2 h mgh mg mv and Bk 22c o s s i n 1 1 . 8 m / s h g h =− = . EVALUATE: For the round trip up the hill and back down, there is negative work done by friction and the speed of the rock when it returns to the bottom of the hill is less than the speed it had when it started up the hill. 7.49. IDENTIFY: Apply Eq.(7.7) to the motion of the stone. SET UP: 1 1 other 2 2 K ++ = + Let point 1 be point A and point 2 be point B . Take 0 y = at point B . EXECUTE: 11 2 , mgy mv mv with 20.0 m h = and 1 10.0 m/s v = 2 21 2 . 2 m / s vvg h =+= EVALUATE: The loss of gravitational potential energy equals the gain of kinetic energy. (b) IDENTIFY: Apply Eq.(7.8) to the motion of the stone from point B to where it comes to rest against the spring. SET UP: Use 1 1 other 2 2 , K with point 1 at B and point 2 where the spring has its maximum compression x . EXECUTE: 12 2 UU K === 2 1 2 K mv = with 1 22.2 m/s v = 2 1 other el k 2 , f W m g s k x with 100 m s x = + The work-energy relation gives 1o t h e r 0. KW 1k 0 mv mgs kx −− = Putting in the numerical values gives 2 29.4 750 0. xx + The positive root to this equation is 16.4 m. x = EVALUATE: Part of the initial mechanical (kinetic) energy is removed by friction work and the rest goes into the potential energy stored in the spring. (c) IDENTIFY and SET UP: Consider the forces. EXECUTE: When the spring is compressed 16.4 m x = the force it exerts on the stone is el 32.8 N. Fk x The maximum possible static friction force is 2 max (0.80)(15.0 kg)(9.80 m/s ) 118 N. g = EVALUATE: The spring force is less than the maximum possible static friction force so the stone remains at rest.
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7-16 Chapter 7 7.50. IDENTIFY: Once the block leaves the top of the hill it moves in projectile motion. Use Eq.(7.14) to relate the speed B v at the bottom of the hill to the speed Top v at the top and the 70 m height of the hill.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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226_PartUniversity Physics Solution - Potential Energy and...

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