Question

Consider the following three cases 1. The finger is pushing but not hard enough. Neither block moves 2. The finger is pushing hard enough that the two blocks are speeding up 3. The blodies have sped up, and are now moving at a constant speed. The finger still has to push to keep them going at a constant velodty. In all cases where the blocks are moving, they are moving together. (Block is not sliding on block A.) Block A has a mass of 0.8 kg; block has a mass of 0.5 kg. the coefficient of friction between the black and the table is 0.2 and the coeffident of friction between the two blods is 0.1. (You may use 9 - 10 N/kg and you may treat kinetic and static friction as the same.) (a) Por case 1, what is the maximum fore the finger can exert without the blocks beginning to move? (b) For case 2, while the blocks are speeding up, the finger is pushing with a force of 5.6 N. Can you find the acceleration of block A? If so, find it and show your work. If not, explain why not. (Show your work or explain why not.) This answer has not been graded yet (c) Find the time At needed for this force to bring the black from rest to a speed of 10 cm/s. (of it is not possible to find the acceleration in part (b), enter NONE.) (d) For case 2, we are assuming that the blocks are moving together. This means that block B is speeding up. But the finger is not touching block . What force is responsible for the acceleration of block B? The force of friction on block A by the table The force of friction on block by block A The force of friction on block A by block B The weight of block A Find Its magnitude and direction. magnitude direction de n Is our assumption correct? Will block B indeed be able to speed up with block A without slipping? Explain. This answer has not been graded yot (e) Suppose the bottom block has a mass of 1.2 kg and the top block has a mass of 0.1 kg What foren do you need to exert to keep the blocks moving at a constant speed of 10 cm/s? EN

Answer 1

o le Griven :- force) 1.39 (force due to your mass of block B le-012 my-0.5kg mass of block A ma=0.8kg coefficient of friction between block & table , le,= 0.2 coefficient of friction between two blocks l = 0.1 Free body agram of combine block (system) Til Total mass = matma = 6:5+0 18 = 1:3lag from F.BD t(friction of T surfaced system is rest etx=of sty zo 1.39 (force due to gravity.) = F = t maxinum & N=1.39 friction force at surface I traxe = 4, N=1639x002 from egno = 1-3 X1 X 02 = 2.6N e to s tmaxt > F e traxe < 2-6N maximum force IF = 2.6N/ 6 Now, F = 5.6 N Here F > 2.6 system Hence blocks will mone suppose both blick move with some acceleration & acceleration of blocks is a. From Newton's and low- 26N L {fp = ma = F-2-6 = 1.39 >> a = 5.6-2.6, 3:6 - 2:307 m/s2 BoD of 39 11:39 1.3 1.3

YOYO check FB:D of blocks since both blick move with same accelercelon fence acxelerohen static force will acts on surface from Newton's and low 0.5g (Force {fy = mga Isty=0 f = 0.5% 2.307 N,=0.5g = 1.155N sus face Here I t = N 4 t o> tmax mex = 0.18 0.5X10 Our aesumption is false = 0.5N Both blocksmove with different acceleration, I FB1) of block a FoBoD of block B t = torax-oost A t=005 & due to maximum friction at CBD of B. Fu= 2067 8.89 0.5g f = mz9B F-16 fu = MAPA => = 5.6-2.6-005 0.8 -2.5 Long 0.8 = 3.125m/s2 @ u=0 &=10cm/s = 0.1mys A = 3-125 m/s2 from v= utat Dol = 0+3-125t = 0-032 sec. 3-125

As well- The force of friction on blockb surface by block A & its magnitude is OSN direction is rightward. Our assumption is not true both Block move with different acceleratius of we take both bleek move together then the static friction cet surface @ becomes greater than the maximum friction cet surface @ Hence our aesumption is not true Total marga109+01= 10312g u=pornts = 1-39 combine FOBD Freeded 1.3! te = 0.281-39 = 206N 1-2 1039 relcity is constant means acceleratian will be zero o Feeded = t = 2-6N APS