PHYSICS 111 HOMEWORK SOLUTION #7

1 PHYSICS 111 HOMEWORKSOLUTION #7 March 10, bead slides without friction around a looptheloop (see figure below).The bead is released from rest at a height h = a) What is its speed at point A (Use the following as necessary:the acceleration due to gravity g, and R.) b)How large is the normal force on the bead at point A if itsmass is grams?a)In the absence of friction, the energy of the bead is conserved so that on eachpoint of the track we have:mv22+mgH=Constant0 +mgh=12mv2A+mg 2R12mv2A=mg( 2R)vA= 2gR( 2)= )On point A, the centripetal force is the sum of gravity force and the normalforce:N+mg=mv2R=m ( 1)= 10 3= objects are connected by a light string passing over a light, fric-tionless pulley as shown in the figure below.

2 The object of massm1= kg is released from rest at a height h = m above the table. a)Using the isolated system model, determine the speed of theobject of massm2= kg just as the object hits thetable. b)Find the maximum height above the table to which the )In the isolated system assumption, the total energy of the whole system isconserved. Let s take the table as our reference for heights. At rest, the totalenergy is just the potential energy due to gravity of massm1that massm1hits the table this energy becomes :m1v22+m2v22+ The two masses are moving with the same speed due to the ,m1gh=12m1v2+12m2v2+m2ghv= 2gh(m1 m2)m1+m2= 2 ( ) + )Massm1still on the table, Massm2will keep ascending till its speed becomeszero where it reaches a height H.

3 Energy is again conserved between the mo-mentsm2was at height h and height H:m2gH=m2gh+m2v22H=h+v22g= +522 coefficient of friction between the block of mass m1 = kgand the surface in the figure below is k= The system startsfrom rest. What is the speed of the ball of mass m2 = kg whenit has fallen a distance h = m?Due to friction, energy is not conserved. We can follow the same steps we usedin problem 3 HOMEWORK # 5 to dervive the acceleration of the two masses:a=g(m2 km1)m1+m2= + can use the time-independent equation :v2f v2i= 2ahvf= 2ah= 2 block is set into motion up an inclined plane with an initialspeed ofvi= m/s (see figure below).

4 The block comes to restafter traveling d = m along the plane, which is inclined at anangle of = to the horizontal. a) For this motion, determine the change in the block s kineticenergy. b)For this motion, determine the change in potential energy ofthe blockEarth system. c) Determine the friction force exerted on the block (assumedto be constant). d) What is the coefficient of kinetic friction?a) +y+x ~N~fkm~ change in kinetic energy is just:12mv2f 12mv2i= 0 )The change in potential energy is just:mghf mghi=mgdsin = 3 sin 30 = )The change in the total energy is the work done by friction forces:Wfriction=~fk ~d= Etotal fk d= Etotalfk= Etotald= + )fk= kN= kmgcos k=fkmgcos = cos 30= child s pogo stick as shown below stores energy in a spring witha force constant of 10 4N/m.

5 At position A (xA=-130m) ,the spring compression is a maximum and the child is momentarilyat rest. At position B , the spring is relaxed and the child is movingupward. At position C, the child is again momentarily at rest at thetop of the combined mass of the child and the pogo stick is kg. Al-though the boy must lean forward to remain balanced, the angle issmall, so let s assume the pogo stick is vertical. Also assume the boydoes not bend his legs during the motion. a)Calculate the total energy of the child-stick-Earth system,taking both gravitational and elastic potential energies as zerofor x = 0. b)DeterminexC c) Determine the value of x for which the kinetic energy of thesystem is a maximum.

6 D) Calculate the child s maximum upward Total Energy will be conservedthroughout the whole )The system is at rest at point A (no kinetic energy), the total energy is justthe potential energy:Etotal=12kx2A+mgxA=12 25500( )2+ ( )= )C is another point of rest and the total energy is just the potential =12kx2C+mgxC=255002x2c+ this equation leads to two values ofxC:x= which is the lowerextermum (Point A), andxC= )Another way to see Energy conservation law is to understand that that what sgained as a potential energy is lost as a kinetic energy and vice versa. Kineticenergy is at its maximum when potential energy is at its +mgxThe minium is to be found by the condition :dEpot(x)dx= 0kx+mg= 0x= mgk= positionx= found above is the point of maximum kineticenergy.

7 Energy conservation will give:Etotal= +Epot=Ekin+mgxEkin= ( )= the maximum upward speed is :v= 2 Ekinm= 2 block is released from rest at point A in the figure track is frictionless except for the portion between points B andC, which has a length of m. The block travels down the track, hitsa spring of force constant 2 200 N/m, and compresses the spring from its equilibrium position before coming to rest the coefficient of kinetic friction between the block and therough surface between points B and choose the horizontal surface as the 0 potential A and B, energy is conserved and it amounts to :mgh=12mv2 BBetween B and C, energy is not conserved due to friction.

8 The change in energyis the work done by friction :12mv2C 12mv2B=~fk ~d= kNd= kmgdRememberN=mgfrom 2nd C and the final position, energy is conserved and it amounts to12mv2C=12k( x)2; xbeing the spring compression in all, kmgd=12mv2C 12mv2B=12k( x)2 mgh k=hd k( x)22mgd=36 2200 10 6= block of massm1= kg is connected to a block of massm2= kg by a massless string that passes over a light, frictionless block is connected to a spring that has negligible massand a force constant of k = 220 N/m as shown in the figure spring is unstretched when the system is as shown in the figure,and the incline is frictionless. The block is pulled a distance h= cm down the incline of angle = and released from the speed of each block when the spring is again s use the isolated system model consisting of the two blocks.

9 Energy con-servation requires that the change in potential energy for the systemm1+m2is of same magnitude as the change in kinetic energy but with oposite sign. Kin= Pot11 The two blocks are connected and move with the same speed v, we should have:12m1v2+12m2v2= (m1g H1+m2g H2+ 0 12k( H2)2)(m1+m2)v2= (m1ghsin m2gh 12kh2)v= 2gh(m2 m1sin ) +kh2m1+m2= 2 (32 18 sin 40) + 220 ( )232 + 18= electric scooter has a battery capable of supplying 105 Wh ofenergy. If friction forces and other losses account for of theenergy usage, what altitude change can a rider achieve when drivingin hilly terrain, if the rider and scooter have a combined weight of930 N?

10 Let s convert the supplied energy to = 105 = 378000 JOnly 40% will be available for the rider:40% 378000 = 151200 JThis amount of energy is then used to reach a height h, the gain in poten-tial energy is provided by the 151200 Joules supplied by the scooter ,mgh= 151200Jh=151200930= stone is held m above the top edge of a water well andthen dropped into it. The well has a depth of m. a)Relative to the configuration with the stone at the top edgeof the well, what is the gravitational potential energy of thestoneEarth system before the stone is released. b)Relative to the configuration with the stone at the top edgeof the well, what is the gravitational potential energy of thestoneEarth when it reaches the bottom of the well?