A heavy block, labeled \"A\", is sitting on a table. On top of that block is a l

Question

A heavy block, labeled "A", is sitting on a table. On top of that block is a lighter block, labeled "B" as shown in the figure below.

(b) For case 2, while the blocks are speeding up, the finger is pushing with a force of 9.0 N. Can you find the acceleration of block A? If so, find it and show your work. (express answer in m/s2 ) If not, explain why not.

(c) Find the time ?t needed for this force to bring the block from rest to a speed of 10 cm/s. (If 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 B. Find its magnitude and direction.

(e) Suppose the bottom block has a mass of 1.4 kg and the top block has a mass of 0.1 kg What force do you need to exert to keep the blocks moving at a constant speed of 10 cm/s? (express answer in Newtons)

Explanation / Answer

a)consider block A and block B as a single system.

then maximum force that be exerted by the finger without moving the blocks is the maximum friction that is exerted by the table

on both the blocks as a single system.

hence maximum force=0.4*(0.9+0.6)*10=6 N

b)as we have assumed that coefficient of both static and kinetic friction are equal,

when blocks are moving , they are experiencing same friction force as static friction i.e. 6 N

hence net force on the blocks=9-6=3 N

then acceleration of the system of blocks (as they are moving together)=3/(0.9+0.6)=2 m/s^2

c)initial veloicty=0

final veloicty=10 cm/sec=0.1 m/s

as we know, final veloicty=initial veloicty+acceleration*time
==>0.1=0+2*t

==>t=0.1/2=0.05 seconds

hence it will take 0.05 seconds for the blocks to accelerate to 10 cm/seconds veloicty.

d)as block B is moving together with block A, it has same acceleration as block A
hence force on block B is towards right direction with a magnitude of 0.6*2=1.2 N

this force is due to the friction between the two blocks.

e)to make the moving at constant speed, you need to exert a force equal to their static friction force.

hence force required=0.4*(1.4+0.1)*10=6 N

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