There are 10 identical blocks lined up in a row. Each block has a mass of 240 g.

Question

There are 10 identical blocks lined up in a row. Each block has a mass of 240 g. A force

F = 4.00 N

is applied to block 1, as shown in the picture. The coefficient of kinetic friction between each block and the horizontal surface is 0.160. At the instant shown in the picture, the blocks are sliding to the left at a speed of 2.20 m/s. Use

g = 10 N/kg

for this problem.

(a) Calculate the magnitude of the net force acting on block 6.
(b) Calculate the magnitude of the force applied on block 6 by block 7.
(c) To answer the questions above, it would be reasonable to assume that the blocks were on a table or on a floor, and that the table/floor was not accelerating. Let's now repeat the scenario, but the blocks are sliding across the floor of an elevator that is accelerating up with an acceleration of 1.70 m/s2. Calculate the magnitude of the horizontal component of the net force acting on block 6 in that case.

Please help I'm really stuck :( The answer for a is NOT 1.74, the answer for b is not 1.37 and the answer for c is not 0.16

Explanation / Answer

a) Lets take all 10 blocks as a system

then friction f = u.Mg

Using Fnet = ma on this system ,

F - f = Ma

4 - 0.160 x (10 x 0.240)x9.81 = (10 x 0.240)a

0.233 = 2.4a

a = 0.097 m/s^2

now using Fnet = ma on block 6 ,

Fnet = 0.240 x 0.097 = 0.0233 N

b) by taking 1 to 6 blocks as a single system ,

Fnet = ma

F - friction force f - force by block 7 = Ma

4 - (0.16 x 6 x 0.240 x 9.81) - N = (6 x 0.240 ) x 0.097

N = 1.6 N

C) Now normal reaction due to table will change so Value of friction will also
change.

Taking all blocks as a single system,

N - mg = ma

N = 10 x 0.240 x (9.81 + 1.70) = 27.62 N

So friction f = 0.16 x 27.62 = 4.42 N

Now value of friction is more than F .

they will start decelerating.

f - F =Ma

4.42 - 4 = (10 x 0.240)a

a = 0.174 m/s^2

No taking 6 block system

N = 6 x 0.240 (9.81 + 1.70) = 16.57

friction = 0.16 x 16.57 = 2.65 N

no balancing forces in horizontal, Fnet = ma

f + due to block7 - F = ma

2.65 + N! - 4 = (6 x 0.240x 0.174)

Ni = 1.60 N

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