Try doing this problem without a calculator. Three blocks are placed side-by-sid

Question

Try doing this problem without a calculator. Three blocks are placed side-by-side on a frictionless horizontal surface and subjected to a horizontal force F, that causes the blocks to accelerate from rest to the right. Consider case 2 only. In case 2, if the horizontal force F is 40.0 N, find the following. Just for fun, think about whether any of your answers would change if we asked the same questions about case 1.

(a) the magnitude of the acceleration of the system

(b) the magnitude of the net force acting on the 2.0-kg block


(c) the magnitude of the force exerted by the 5.0-kg block on the 2.0-kg block

Explanation / Answer

Therefore the acceleration of the system F = ma

so a = F/m = 40/(2+5+3)

a = 4 m/s^2

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Now from equation, FNet = ma

Fnet = 2* 4 = 8 m/s^2

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F- f = 2*4 = 8 m/s^2

f = 40 -8 = 32 N

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just for fun in case of 1:

1. a = F/m = 40/10 = 4 m/s^2

2. Fnet = 2* 4 = 8 m/s^2

3. F3 on 3kg = 3*4 = 12 m/s^2

Force on 2kg = F- F3 = ma = 2 * 4

F = 8 + 12 = 20 N

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