exerted on block 2 by block 1. We have quite a few pieces of information to include. First, Newton s second law for blocks 1 and 2: r ( ) = = =

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1 8.8. Model: Blocks and are our systems of interest and will be treated as particles. Assume a frictionless rope and massless pulley. Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration constraint is a = a = a, where a will have a positive value. There are two real action/reaction pairs. The two tension forces will act as if they are action/reaction pairs because we are assuming a massless rope and a frictionless pulley. Make sure you understand why the friction forces point in the directions shown in the free-body diagrams, especially force f r eerted on block by block. We have quite a few pieces of information to include. First, Newton s second law for blocks and : r F f T µ n T m a m a F n m g 0 n m g ( net on ) = = k = = ( ) = = = net on N y ( ) = = = = ( Fnet on ) = n n mg = 0 N n = n + mg F T f f T T f µ n T m a m a net on pull pull k y We ve already used the kinetic friction model in both -equations. Net, Newton s third law: n = n = m g f = f = µ kn = µ k mg T = T = T Knowing n, we can now use the y-equation of block to find n. Substitute all these pieces into the two -equations, and we end up with two equations in two unknowns: ( ) = µ k mg T= ma Tpull T µ kmg µ k m + m g ma Subtract the first equation from the second to get T µ ( 3m + ) Tpull µ k ( 3m + m ) g = ( m + m ) m g a a = =.77 m/s m + m pull k

2 8.34. Model: Use the particle model for the block of mass M and the two massless pulleys. Additionally, the rope is massless and the pulleys are frictionless. The block is kept in place by an applied force r F. Solve: Since there is no friction on the pulleys, T = T 3 and T = T 5. Newton s second law for mass M is T w = 0 N T = Mg = ( 0. kg)( 9.8 m/s )= 00 N Newton s second law for the small pulley is T T + T3 T = 0 N T = T3 = = 50 N = T5 = F Newton s second law for the large pulley is T T T T = 0 N T = T + T + T = 50 N

3 8.38. Model: Assume the particle model for the book (B) and the coffee cup (C), the models of kinetic and static friction, and the constant-acceleration kinematic equations. Solve: (a) Using v = v0 + a( 0), = ( ) + ( ) 0 m /s 3.0 m/s a a = 4.5 m /s To find, we must first find a. Newton s second law for the book and the coffee cup is Fon B nb wb cos0 0 N nb.0 kg 9.8 m/s cos0 9. N y ( ) = = = ( )( ) = ( ) = = Fon B T fk wbsin 0 mbab F T w m a The last two equations can be rewritten, using a C = a B = a, as Adding the two equations, ( ) = = on C y C C C T n m gsin0 = m a T m g = m a µ k B B B am ( + C m )= B gm + C m sin0 µ 9. N B k ( ) ( ) (.5 kg) = ( 9.8 m/s )[ 0.5 kg + (.0 kg) ] ( 0.0)( 9. N) a sin 0 a = 6.73 m/s Using this value for a, we can now find as follows: = 4.5 m /s = 4.5 m /s a 673. m/s C = m C

4 (b) The maimum static friction force is fs ma = µ snb = ( 050. )( 9. N )= 460. N. We ll see if the force f s needed to keep the book in place is larger or smaller than f s ma. When the cup is at rest, the string tension is T = m C g. Newton s first law for the book is ( Fon B) = fs T wb sin0 = fs mcg mbgsin0 = 0 Because f > f the book slides back down. s s ma, ( ) f = M + M sin0 g = 8.5 N s C B

5 8.40. Model: Use the particle model for the two blocks. Assume a massless rope, and massless, frictionless pulleys. Note that for every meter block moves forward, one meter is provided to block. So each rope on m has to be lengthened by one-half meter. Thus the acceleration constraint is a = a. Solve: Newton s second law for block is T = m a. Newton s second law for block is T w = m a. Combining these two equations gives mg ( ma ) mg = m( a ) a[ 4m + m]= mg a = 4m + m where we have used a = a. Assess: If m = 0 kg, then a = g. This is what is epected for a freely falling object.

6 9.8. Model: Use the particle model for the sled, the model of kinetic friction, and the impulse-momentum theorem. Note that the force of kinetic friction f k imparts a negative impulse to the sled. Solve: Using p = J, we have tf tf pf pi = F () t dt = fk dt = fk t mvf mvi = µ kn t = µ kmg t ti ti We have used the model of kinetic friction f k = µ k n, where µ k is the coefficient of kinetic friction and n is the normal (contact) force by the surface. The force of kinetic friction is independent of time and was therefore taken out of the impulse integral. Thus, t g v v = ( i f ) = µ k ( )( ( 8.0 m/s 5.0 m/s) =. s m/s )

7 9.4. Model: Choose car + rainwater to be the system. There are no eternal horizontal forces on the car + water system, so the horizontal momentum is conserved. Solve: Conservation of momentum is p f = p i. Hence, ( mcar + mwater )( 0 m/s)= ( mcar )( m/s)+ ( mwater )( 0 m/s) ( water )( )= ( )( ) 5,000 kg + m 0 m/s 5,000 kg m/s = 500 kg m water

8 9.0. Model: Because of eternal friction and drag forces, the car and the blob of sticky clay are not eactly an isolated system. But during the collision, friction and drag are not going to be significant. The momentum of the system will be conserved in the collision, within the impulse approimation. Solve: The conservation of momentum equation p f = p i is ( m + m )( v ) = m ( v ) + m ( v ) C B f B i B C i C 0 kg m/s 0 kg 500 kg (.0 m/s ) ( v ) = 300 m/s = ( )( v ) + ( ) i B i B Assess: This speed of the blob is around 600 mph which is very large. However, we must point out that a very large speed is epected in order to stop a car with only 0 kg of clay.

9 9.4. Model: We will model the two fragments of the rocket after the eplosion as particles. We assume the eplosion separates the two parts in a vertical manner. This is a three-part problem. In the first part, we will use kinematic equations to find the vertical position where the rocket breaks into two pieces. In the second part, we will apply conservation of momentum to the system (that is, the two fragments) in the eplosion. In the third part, we will again use kinematic equations to find the velocity of the heavier fragment just after the eplosion. Solve: The rocket accelerates for.0 s from rest, so ( )= +( ) vy = v0y + ay t t0 0 m/s 0 m/s ( s 0 s )= 0 m/s y = y0 + v0y( t t0)+ ay t t At the eplosion the equation p fy = p iy is m v m v m m v y H y H H y ( 0) = ( 0 )( ) = m m m / s s 0 m ( ) + ( ) = ( + ) ( 500 kg)( vy) + ( 000 kg)( vy) = ( 500 kg)( 0 m / s) To find (v y ) H we must first find (v y ), the velocity after the eplosion of the upper section. Using kinematics, ( 3y) = ( y) + (. )( 3 ) ( v y ) = ( ) v v 9 8 m/s y y Now, going back to the momentum conservation equation we get ( 500 kg )( m / s)+ ( 000 kg)( ) = ( 500 kg )( 0 m / s) The negative sign indicates downward motion. H H ( ) = 98. m / s 530 m 0 m m / s. v y ( ) = v y 0. 0 m/s H