Consider the following three cases. 1. The finger is pushing but not hard enough
ID: 1784169 • Letter: C
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 blocks have sped up, and are now moving at a constant speed. The finger sl has to push to keep them going at a constant velocity. In all cases where the blocks are moving, they are moving together. (Block B is not sliding on block A.) Block A has a mass of 0.6 kg; block B has a mass of 0.1 kg. the coefficient of friction between the block and the table is 0.3 and the coefficient of friction between the two blocks is 0.2. (You may use g = 10 N/kg and you may treat kinetic and static friction as the same. (a) For case 1, what is the maximum force 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 4.1 N. Can you find the acceleration of block A? If so, find it and show your work. If not, explain why not.Explanation / Answer
a)
M = total mass of A and B = 0.6 + 0.1 = 0.7 kg
us = coefficient of static friction = 0.3
maximum force without the blocks being moved is given as
F = static frictional force = us Mg
F = (0.3) (0.7) (9.8)
F = 2.1 N
b)
fk = kinetic frictional force = uk Mg = (0.3) (0.7) (9.8) = 2.1 N
F = applied force = 4.1 N
a = acceleration
force equation is given as
F - fk = M a
4.1 - 2.1 = (0.7) a
a = 2.86 m/s2
c)
a = 2.86 m/s2
Vi = initial velocity = 0 m/s
Vf = final velocity = 10 cm/s = 0.1 m/s
t = time taken
using the equation
a = (Vf - Vi )/t
2.86 = (0.1 - 0)/t
t = 0.035 sec
d)
the force of friction on block B by block A
f = magnitue of force on B = mb a = 0.1 x 2.86 = 0.286 N
direction : towards right