11.8. Model: Work Done by a Force on a Particle Is Defined As Where Is the Particle

Physics I

Homework XI CJ Chapter 11: 8, 14, 16, 32, 40, 46, 54

     11.8. Model: Work done by a force F on a particle is defined as W F   r , where r is the particle’s displacement.

Visualize:

Solve: (a) The work done by gravity is

  垐 2 Wg  w   r (  m g j ) N  ( 2 . 2 5  0 . 7 5 ) j m   ( 2 . 0 k g ) ( 9 . 8 m / s ) ( 1 . 5 0 m ) J   2 9 . 4 J   (b) The work done by hand is WH F h a n d o n b o o k   r . As long as the book does not accelerate, 11.14. Model: Use the work-kinetic energy theorem. Visualize: Please refer to Figure Ex11.14. Solve: The work-kinetc energy theorem is

1 1 x f K  m v2  m v 2  W  F d x  a r e a u n d e r t h e f o r c e c u r v e f r o m x t o x 2f 2 i x i f x i 1 1 1 x f m v2 ( 2 k g ) ( 4 . 0 m / s ) 2  m v 2  1 6 J  F d x  a r e a f r o m 0 t o x f f  x 2 2 2 0 m 1 1 A t x 2 m : ( 2 . 0 k g ) v2  1 6 J  (  1 0 N ) ( 2 m )   1 0 J  v  2 . 4 5 m / s 2f 2 f 1 A t x 4 m : ( 2 . 0 k g ) v2  1 6 J  0 J  v  4 . 0 m / s 2 f f

11.16. Model: Use the definition Fs   d U/ d s . Visualize: Please refer to Figure Ex11.16. Solve: Fx is the negative of the slope of the potential energy graph at position x. Between x  0 cm and x  10 cm the slope is

slope  (Uf U i ) / ( x f  x i )  ( 0 J  1 0 J ) / ( 0 . 1 m  0 . 0 m )   1 0 0 N

Thus, Fx  100 N at x  5 cm. The slope between x  10 cm and x  20 cm is zero, so Fx  0 N at x  15 cm. Between 20 cm and 40 cm, slope  ( 1 0 J 0 J ) / ( 0 . 4 0 m  0 . 2 0 m )  5 0 N

At x  25 cm and x  35 cm, therefore, Fx  50 N. 11.32. Solve: (a) A kilowatt hour is a kilowatt multiplied by 3600 seconds. It has the dimensions of energy. (b) One kilowatt hour is energy 1 k w h ( 1 0 0 0 J / s ) ( 3 6 0 0 s )  3 . 6  1 06 J Thus 3 . 6 1 06 J  5 0 0 k w h  ( 5 0 0 k w h )   1 . 8  1 09 J 1 k w h 

11.40. Model: Model the rock as a particle, and apply the work-kinetic energy theorem. Visualize:

Solve: (a) The work done by Bob on the rock is 1 1 1 1 W  K  m v2  m v 2  m v 2 ( 0 . 5 0 k g ) ( 3 0 m / s ) 2  2 2 5 J B o b2 1 2 0 2 1 2

(b) For a constant force, WB o b F B o b  x  F B o b  W B o b /  x  2 2 5 N .

(c) Bob’s power output is PB o b F B o b v r o c k and will be a maximum when the rock has maximum speed. This is just as he releases the rock with vrock  v1  30 m/s. Thus, Pm a x F B o b v 1 6 7 5 0 W  6 . 7 5 k W .

11.46. Model: Identify the truck and the loose gravel as the system. We need the gravel inside the system because friction increases the temperature of the truck and the gravel. We will also use the model of kinetic friction and the conservation of energy equation. Visualize:

We place the origin of our coordinate system at the base of the ramp in such a way that the x-axis is along the ramp and the y-axis is vertical so that we can calculate potential energy. The free-body diagram of forces on the truck is shown.

Solve: The conservation of energy equation is K1 U g 1   E t h  K 0  U g 0  W e x t . In the present case, Wext  0 J, v1x  0 m/s, Ug0  0 J, v0x  35 m/s. The thermal energy created by friction is

Et h ( f k ) ( x 1  x 0 )  ( k n ) ( x 1  x 0 )   k ( m g c o s 6 . 0  ) ( x 1  x 0 ) 2 ( 0 . 4 0 ) ( 1 5 , 0 0 0 k g ) ( 9 . 8 m / s ) ( c o s 6 . 0  ) (x1  x 0 )  ( 5 8 , 4 7 8 J / m ) ( x 1  x 0 ) Thus, the energy conservation equation simplifies to 1 0 Jm g y  ( 5 8 , 4 7 8 J / m ) ( x  x )  m v 2  0 J  0 J 1 1 02 0 x 1 ( 1 5 , 0 0 0 k g ) ( 9 . 8 m / s2 ) [ (x x ) s i n 6 . 0  ]  ( 5 8 , 4 7 8 J / m ) ( x  x )  ( 1 5 0 0 0 k g ) ( 3 5 m / s ) 2 1 0 1 0 2

(x1  x 0 )  1 2 4 . 4 m Assess: A length of 124.4 m at a slope of 6º seems reasonable.

11.54. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the physics student as a particle and apply the law of conservation of energy. Our system is comprised of the spring, the student, and the ground. We also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the end of the compressed spring that is in contact with the student.

Solve: (a) The energy conservation equation is

K1 U g 1  U s 1   E t h  K 0  U g 0  U s 0  W e x t 1 1 1 1 m v2 m g y  k( x  x ) 2  0 J  m v 2  m g y  k ( x  x ) 2  0 J 21 1 2 1 e 2 0 0 2 1 0

Since y1  y0  10 m, x1  xe, v0  0 m/s, k  80,000 N/m, m  100 kg, and (x1  x0)  0.5 m, 1 1 k m v2 k( x  x ) 2  v  ( x  x )  1 4 . 1 4 m / s 21 2 1 0 1m 1 0 (b) The work of friction creates thermal energy. Applying the conservation of energy equation once again:

K2 U g 2  U s 2   E t h  K 0  U g 0  U s 0  W e x t 1 1 m v2 m g y 0 J  f  s  0 J  m g y  k ( x  x ) 2  0 J 22 2 k 0 2 1 0

With v2  0 m/s and y2  (s)sin30º, the above equation is simplified to 1 m g( s ) s i n 3 0   n  s  m g y  k ( x  x ) 2 k 02 1 0 From the free-body diagram for the physics student, we see that n w c o s 3 0  . Thus, the conservation of energy equation gives 1 s( m g s i n 3 0   m g c o s 3 0  )  m g y  k ( x  x ) 2 k 02 1 0

Using m  100 kg, k  80,000 N/m, (x1  x0)  0.50 m, y0  10 m, and k  0.15, we get

1 2 m g y0 k( x 1  x 0 ) s 2  3 2 . 1 m m g ( s i n 3 0  k c o s 3 0  )

Assess: y2 (  s ) s i n 3 0   1 6 . 0 5 m , which is greater than y0  10 m. The higher value is due to the transforma- tion of the spring energy into gravitational potential energy.