<p> Solutions to Mid-Term Exam for GP I, MATH, Fall 2011. 1. 20% You are lowering the boxes down a ramp as shown in figure. Both boxes move with a constant speed of 15.0 m/s. The coefficient of kinetic friction between the lower box and the ramp is 0.444. The coefficient of static friction between the upper and lower boxes is 0.800.</p><p>(a) What force do you need to exert to accomplish this? (b) What are the magnitude and direction of the friction force on the upper box? Solution</p><p>Apply Fnet = m a to the composite object consisting of the two boxes and to the top box. The friction the ramp exerts on the lower box is kinetic friction. The upper box doesn’t slip relative to the lower box, so the friction between the two boxes is static. Since the speed is constant the acceleration is zero.</p><p>(a) </p><p> fk+ T - m tot gsinf = mtot a = 0</p><p> ntot= m tot g cosf fk= m k m tot g cos f</p><p> T= mtot g (sinf - m k cos f) 骣 2.50 4.75 =琪 -0.444创 ( 32.0 + 48.0 kg) 9.80 m/s2 琪 2 2 2 2 ( ) 桫 2.50+ 4.75 2.50 + 4.75 = 57.1 N (b)</p><p>2 2.50 fs = mg sinf =( 32.0 kg) 创( 9.80 m/s) = 146 N 2.502+ 4.75 2</p><p>1. 15% Each seat of the “Giant Swing” is suspended from a cable 5.00 m long attached to a horizontal arm 3.00 m long (see figure).</p><p>Find the time of one revolution of the swing if the supporting cable makes a 30 angle with the vertical.</p><p>Solution The person moves in a circle of radius</p><p>R =3.00 m + ( 5.00 m) � sin 30.0 5.50 m</p><p>Let F be the force applied to the seat by the rod. v2 x: Fsin 30.0 = m R y: Fcos30.0 = mg</p><p> v= Rg tanq =( 5.50 m) 创( 9.80 m/s2 ) tan30.0 = 5.58 m/s</p><p>2p R 2p ( 5.50 m) T = = = 6.19 s v 5.58 m/s</p><p>3. 30% The figure shows an accelerometer. (a) How is  related to the acceleration of the system.</p><p>(b) If you can vary m1 and m2, what is the largest angle  you can achieve?</p><p>Solution</p><p>(a) Ball: x: Tsinq = m a y: Tcosq = m g a  tanq = g (b)</p><p>For m2 :</p><p> x: T= m2 a</p><p>For m1 :</p><p> y: m1 g- T = m 1 a</p><p>骣 m  a= 琪 1 g 桫m1+ m 2</p><p> m tanq = 1 m1+ m 2</p><p>The largest  occurs when m1? m 2 so that</p><p> tanq 1 or q  45 4. 15%</p><p>A truck of mass m has brake failure and careens down an icy slope of angle </p><p> and negligible friction ( see figure ). The initial speed of the truck is v0 , and after careening a distance L down the slope, it climbs onto a truck ramp of angle </p><p> and rolling coefficient r. What is the distance that the truck moves up the ramp before it stops? Solution 1 Initially, E= mv2 + mgL sina . 02 0 1 At the bottom of the slope: E= mv2 1 2</p><p>When the truck stops: E2 = m g d sin b Let d be the distance the truck moves up the ramp.</p><p>Work done by friction: Wf= m r ( m gcos b ) d</p><p>Energy conservation: E1= E 0</p><p>Energy-work theorem: E1- E 2 = W f</p><p>1  mv2 + mgLsina - m g d sin b = m( m g cos b ) d 2 0 r 1 v2 + L sina 2g 0 d = sinb+ mr cos b</p><p>5. 20% When an object is in a circular orbit of radius r around the Earth, the period of the orbit is T, and the orbital speed v. Show that when the object is moved into a circular orbit of slightly larger radius r + r, where r << r, its new period is T + T, where 3p Dr p Dr DT = Dv = - v T Solution</p><p> v2 GM GM = E  v = E r r2 r</p><p>2p r 2p r3 2 T = = v GM E</p><p>3 2 3 2 2π 3 2 2πr 骣 D r  T+ D T =( r + D r ) =琪1 + GM E GM E 桫 r</p><p>2πr 3 2骣 3D rπr r 3 1/2 D � 琪1= + T GME桫 2r GM E</p><p>3πr 1/2D r DT = 3p Dr v GM E</p><p>-1 2 v+ D v = GM ( r + D r )-1 2 = GM r - 1 2 1 + Dr E E ( r )</p><p>-1 2 Dr Dr p Dr � GM r 1 =v - GM E 3/2 =v - E ( 2r ) 2 r T p Dr  Dv = - T</p>