Physics 201 Homework - PCC

1 Physics 201 Homework 5 Feb 6, (non-conservative) force propelling a 1500-kilogram car up a 106joulesroad does 106joules of work on the car. The car starts from rest at sea leveland has a speed of m/s at an altitude of 200 meters above sea level. Obtainthe work done on the car by the combined forces of friction and air resistance,both of which are non-conservative propelling force has the effect of increasing the energy level of the car. Thisenergy goes three places: kinetic energy (or speed), potential energy (or altitude),and some is lost to the nonconservative friction. We know the energy going inand we can calculate the amount that goes into kinetic and potential is left must have been lost to the non-conservative forces.

2 In symbols:Wadded Wlost= E= KE+ PEWe know the kinetic energy gained:KE=12mv2=12(1500)( )2= 106 And the potential energy gained:PE=mgh= (1500)( )(200) = 106 Plugging these and the energy added we have:( 106) Wlost= 106+ 106= Wlost= 106It s a technicality, but the question asks for the workdoneby the resistive we have calculated the work lost, we need to flip the sign to emphasize thatthe energy level is decreasing as a consequence of the person pushes a shopping cart at a constant velocity for a(a) newtons(b) 1060 joules(c) -1060 joules(d) nonedistance of meters. She pushes in a direction below the frictional force opposes the motion of the cart.

3 (a) What is themagnitude of the force that the shopper exerts? Determine the work done by (b)the pushing force, (c) the frictional force, and (d) the gravitational (a) Since the cart is moving at constant velocity, the system is in equilibrium,so the net force is zero. In this problem there are four forces involved. Thepushing force, the weight of the cart, the normal support force, and the kineticfriction. The motion is horizontal, so we should align the coordinate that components of the net force are:Fnet,x=Px FFnet,y=Py W+NBoth of these components sum to zero. Since we are given the friction force( newtons), we know that thex-component of the pushing force is also Since we know that the canonical angle of the push is 331 , we have:Px=Pcos 331 = P= (b) The work done by any force is the product of the displacement and thecomponent of the force parallel to the displacement.

4 In this case,x= metersandPx= newtons. Thus,W=xPx= ( )( ) = 1056(c) Similarly for the friction, but this force opposes the motion so its componentis negative:W=xFx= ( )( ) = 1056(d) The gravitational force has no component in thexdirection. Thus,W=xWx= ( )(0) = arrow is fired horizontally. The bowstring exerts an average39 m/sforce of 65 newtons on the arrow over a distance of meters. With what speeddoes the arrow leave the bow?SolutionThe work done by the bowstring isW=xFx= ( )(65) = work has the effect of increasing the energy of the arrow. In this case itincreases the kinetic energy of the arrow. The initial kinetic energy is zero, soKEf= the formula for kinetic energy,KE=12mv2, we can derive the final resultingspeed:12( )(v)2= = v= extreme skier, starting from rest, coasts down a mountain slope that m/san angle of with the horizontal.

5 The coefficient of kinetic friction betweenher skis and the snow is She coasts down a distance of meters beforecoming to the edge of a cliff. Without slowing down, she skis off the cliff andlands downhill at a point whose vertical distance is meters below the fast is she going just before she lands?SolutionThere are two phases to the motion in this problem. The first coasting phase, isbasically an inclined plane problem from Chapter 4. At the end of this phase theskier will have a certain velocity angled down 25 from the horizontal. The secondphase is standard projectile problem which we could solve using the techniquesfrom Chapter 3. However, this since we are only asked about the speed, we coulduse kinetic energy to answer the question.

6 Either approach will work. But sincethis is a Chapter 6 problem, it seems appropriate to use the first phase, there are three forces at work: weight, support, and coordinates aligned with the motion (assume the slope is to the right), thecanonical angles associated with these forces are 295 , 90 , and 180 , magnitudes of these forces areW=mg,N, andFk= ( )(N), respec-tively. We don t knowm, but we will assume it cancels out. We don t knowN,but it must counter-balance they-component of the weight. Thus,N+Wy= ,N= Wy= Wsin = ( )(m) sin 295 = ( )(m)Now there is motion in thex-direction. According to Newton s second law wehaveFnet=ma. The net force is in thex-direction, but friction points in thenegative direction, soFnet,x=Wx Fk= (m)( ) cos 295 ( )( )(m) = ( )(m)2So, the acceleration of the skier down the slope is we have a constant acceleration problem.

7 The quantities area= 0v= ?x= equation to use isv2=v20+ 2ax. Thus,(v)2= (0)2+ (2)( )( ) = v= is the endpoint of the first phase of motion, but it is also the beginning ofthe second phase. We decided to use energy considerations to work this know that she is meters above her landing point. This means she has acertain amout of potential energy:PE=mgh= (m)( )( ) = ( )(m)And her initial kinetic energy isKE=12mv2=12(m)( )2= ( )(m)Since no energy is lost, the potential energy is converted into additional kineticenergy at the end of the drop. So the final kinetic energy isKE= ( )(m) + ( )(m) = ( )(m)Using the formula for kinetic energy,KE=12mv2, we can determine her finalspeed:( )(m) =12(m)(v)2= v= Rocket man has a propulsion unit strapped to his back.

8 He starts from rest23 kJon the ground, fires the unit, and is propelled straight upward. At a height of16 meters, his speed is m/s. His mass, including the propulsion unit, has theapproximately constant value of 136 kilograms. Find the work done by the forcegenerated by the propulsion gravitational force is conservative, so we can say thatW= E= KE+ PENow, initially Rocket man has no kinetic or potential energy because he is atrest and on the ground. But at the end point, he has both kinetic and potentialenergy. Thus,KE=12mv2=12(136)( )2= 1700andPE=mgh= (136)( )(16) = 21325 Now, his initial mechanical energy is zero. But the final mechanical energy isE=KE+PE= 1700 + 21325 = 23025 The work done by the propulsion unit is where this energy came minutes, a ski lift raises four skiers at constant speed to a height of3000 watts140 meters.

9 The average mass of each skier is 65 kilograms. What is the averagepower provided by the tension in the cable pulling the lift?Solution3 Since the work done by the lift has the effect of increasing the potential energyof the skiers, we know by the work-energy theorem thatW= PE=mgh= (4 65)( )(140) = 356720 Since this quantity of work is done over a minute time period (or 120 seconds),by definition, the power involved isP=Work t=356720120= small lead ball, attached to a meter rope, is being whirled in a circle42 metersthat lies in the vertical plane. The ball is whirled at a constant rate of threerevolutions per second and is released on the upward part of the circular motionwhen it is meters above the ground.

10 The fall travels straight upward. In theabsence of air resistance, to what maximum height above the ground does theball rise?SolutionAt the top of its trajectory, the kinetic energy is momentarily zero all the energyis in the gravitational potential energymgh. From this we can calculate theheight. Since we know the initial height, all we need to determine is the ball moves in uniform circular motion with a radius of meters. Eachrevolution is 2 r= meters. The speed is three times this value per second:v= (3)(2 )( ) = , the energy of the ball isKE =12mv2=12(m)( )2= =mgh= (m)( )( ) = a total of ( )(m) joules of energy. Therefore the maximum height mustbePE =mgh= (m)( )(h) = ( )(m) = h= pendulum consists of a small object hanging from the ceiling at the end43 of a string of negligible mass.