Question

A 2-kg block is placed on the top of a 5-kg block The coefficientof kinetic friction between the 5-kg block and the surface isμk =0.2. A horizontal force F is appliedto the 5-kgblock.

a) Draw a free-body diagram for each block. What force acceleratesthe 2-kg block?

b) Calculate the magnitude of the force necessary to pull bothblocks to the right with
acceleration 3 m/s2.

c) Find the minimum coefficient of static friction μs betweenthe blocks such
that the 2-kg block does not slip under an acceleration of 3m/s2?

Answer 1

here's a few hints, that might help get you started. Justkeep in mind that you need to draw all of the external forces onthe diagrams, and thatnewton says that NET force equals m*a.meaning m*a is the difference between the pulling force and thefrictional force. Fnet=F-f=ma


A)to accelerate the blocks the net horizontal force on the bottomblock must be greater than zero. Thus, F>f. Theforce that will acceleratethe 2 kg block will come from the netexternal force applied to the block on the bottom.

example to accelerate a 7 kg block at 3 m.s2
Fnet=ma=7kg*3m/s.2
Fnet=21N
we need to apply a net force of 21 newtons to accelerate theblocks, meaning 21 N more than the frictional resistive forcesf.
Fnet=F-f
f=un=.2*7*9.8=13.72N
21=F-13.72
F= 21+13.72=34.72 N

hint for (C) The top block does not move relative to the bottomone, so the net force on it should be zero.
find the force from the bottom block that causes the system to goin motion, there should be enough friction between the two blocksto keep these fromslipping (f=F such that netF=0)

Answer 2

I think question a) is
F=ma_Fg
F= 2 x 9.8 x 0.2=3.92N
F/m=a
so 3.92/2=1.96ms²
Again, I think

Answer 3

c)For the static friction co-efficient(Us) Consider the 5kg block and all it's forces For the X dir We have fk(for the surface and 5kg) F(pulling foward) fs(the static friction of the top 2kg that sits on the 5kg...this pulls it back) For y dir We have N = Fg2 + Fg5 We have enough information to solve for Us X dir Fnet= ma F -fk -fs = m.a 34.72 - 13.72 -fs = (5)(3) Tidy up the equation to get fs = 6 (Us)(N) = 6 Remember that N for 2kg not the 5kg so N = 2 x 9.8 19.6Us = 6 Us = 0.306 Also for a) The force that accelerates the 2kg block Is the force of friction between the 5kg and the 2kg

Answer 4

Sorry for the above answered layout...the spacing kinda sucks but if you write it out you should be able to understand it