Solutions to practice Exam II solved on the blackboard in class. 1

Solutions to practice Exam II solved on the blackboard in class. 1) The free body diagram of the problem is the following: We write Newton’s law in the horizontal and vertical directions:
ΣFy=may ΣFx=max à μsN=mv2/r à N-­‐mg=0 à N=mg. à v2= μsmgr/m= μsgr à v=10.1 m/s Answer: A. 2) We apply Newton’s law in the x direction for the two blocks separately. Notice that, in the case of block 2, we have rotated the coordinate system such that now the positive x direction is vertical, pointing downwards. We have: T=mAax for block A. mBg-­‐T=mBax for block B. Adding the two equations we have: mBg=(mA+mB)ax from which we can calculate the acceleration of the system: ax =mBg/(mA+mB)=2.00kg*9.81m/s2/(3.00kg+2.00kg)=3.92 m/s2. We then use the kinematic equations to find the time it takes for block B to travel 80.0 cm (0.80 m): x=x0+v0xt+1/2axt2 à x-­‐x0=1/2axt2 à t=0.639 s. Answer: C 3) We use Hooke’s law: F=-­‐kx=-­‐mg à x=mg/k=0.0196 m. Answer: B 4) The free body diagram of the problem is as follows: We write Newton’s law along x and y: ΣFy=may à N-­‐mgCos15°+FSin30°=0 à N=mgCos15°-­‐FSin30° =323.79 N ΣFx=max à mgSin15°-­‐ FCos30°-­‐ μkN = max à ax=(mgSin15°-­‐ FCos30°+ μkN)/m=-­‐1.36m/s2= 1. 36m/s2 upwards. Answer: A. 5) The desk slides at constant speed: it means that the force applied by the student is equal and opposite to the kinetic frictional force. The free body diagram of the problem is the following: We write Newton’s law along x and y: ΣFy=may à N-­‐mg=0 ΣFx=max à F-­‐ μkN = 0 à F= μkN= μkmg à N=mg The work done by the student is given by the force times the distance (4 m): W=Fx= μkmgx=1.26 kJ. Answer: D. 6) The mechanical energy is not conserved in this case. The difference between the final and initial mechanical energy is the work done by friction. The height of the plane is h=10 m. At the top of the plane, the object has a potential energy Ui=mgh=392.4J. The kinetic energy is =0. Therefore, the mechanical energy is Ei=Ui+Ki=Ui. At the bottom of the plane, the kinetic energy is: Kf=1/2mv2=200J. The potential energy is =0. Therefore, the mechanical energy is Ef=Uf+Kf=Kf. The work done by friction is: Wnc=Ef-­‐Ei=-­‐192J. Answer: A. 7) The pendulum is released from a height h=0.25 m. Initially it has a potential energy Ui=mpgh. The kinetic energy is =0. Therefore, the mechanical energy is Ei=Ui+Ki=Ui. At the lowest point, the kinetic energy is: Kf=1/2mpvp2. The potential energy is =0. Therefore, the mechanical energy is Ef=Uf+Kf=Kf. In this case there is no nonconservative force: the mechanical energy is conserved. Kf=Ui à 1/2mpvp2=mpgh à vp2=2gh à vp=2.21 m/s. The momentum of the pendulum as it hits the block is: pp=mpvp=2.21 kg*m/s. It is entirely transferred to the block: pb=pp à mpvp= mbvb à vb=4.42 m/s. Answer: D. 8) The potential energy at point A is transformed into kinetic energy at point B: the mechanical energy is conserved. We have: 1/2mv2=mgh à K=2.0*10-­‐2J Answer: B. 9) The mechanical energy of the system is conserved. At point A, the object only has kinetic energy. At point B, the object only has potential energy. The two have to be equal to each other: Ei=Ui+Ki=Ki. Ef=Uf+Kf=Uf. Ei=Ef à 1/2mv2=mgh à h=1/2v2/g=80m. Answer: D. 10) In this case, the mechanical energy is only conserved along the inclined plane but not along the horizontal surface. The initial mechanical energy is the potential energy: Ei=Ui+Ki=Ui=mgh The work done by the frictional force along the horizontal plane is Wnc=-­‐μkNd where d is the distance from the bottom of the inclined plane where the object stops. When the object stops, both its potential and kinetic energy are =0: Ef=0. Wnc=Ef-­‐Ei à -­‐μkNd=-­‐mgh The normal force on the horizontal surface is equal to the weight of the object: N=mg. We find: -­‐μkmgd=-­‐mgh à μkd=h à d= h/μk=20 m. Answer: D. 11) The impulse is simply the change in the momentum of the ball: I=Δp=pf-­‐pi=m(vf-­‐vi)=0.140kg*(30.0m/s-­‐(-­‐40.0m/s))=9.80kg*m/s=9.80 Ns. Answer: A 12) Answer: C. 13) In this case, the mechanical energy is not conserved. The initial mechanical energy is the potential energy: Ei=Ui+Ki=Ki=1/2mv2. The work done by the frictional force along the horizontal surface is Wnc=-­‐μkNd where is the distance from the bottom of the inclined plane where the object stops. When the object stops, both its potential and kinetic energy are =0: Ef=0. Wnc=Ef-­‐Ei à -­‐μkNd=-­‐Ei The normal force on the horizontal surface is equal to the weight of the object: N=mg. We find: -­‐μkmgd=-­‐Ei à μkmgd=Ei à d= Ei/(μkmg)=1m=100cm. Answer: E. 14) Inelastic collision: the momentum is conserved but not the kinetic energy. The initial momentum is given by the one of the first cart alone (the second one is stationary, therefore its momentum is zero). pi=m1v1. After the collision, the two carts stick together. The final momentum is: pf=(m1+m2)vf. The momentum is conserved: pi= pf à m1v1=(m1+m2)vf à vf= m1v1/(m1+m2)=3.0 m/s. Answer: A.