The 1.0 kg block is tied to the wall with a rope. It sits ontop of the 2.0 hg bl

Question

The 1.0 kg block is tied to the wall with a rope. It sits ontop of the 2.0 hg block. The lowe block is pulled to the rightwitha tension force of 20 N. The coefficient of kinetic friction atboth lower and upper surfaces of the 2.0 kg block is k =40 a) what is the tension in the rope holding the 1.0 kg block tothe wall? b) What is the acceleration of the block 2.0 kg block?

Explanation / Answer

I assume you meant k = 0.40. (a) We can determine the tension of the rope tied to the wall as: T = fk = kmg = 0.4(1kg)(9.8m/s/s) =3.92N (b) Now, for this part, we have the block on top (which we will denoteas A) pushing down on the box (B). So, we can set up the force asfollows: F = fA - fB => F -kmAa - k(mA+ mB) = mBa So we get: a = [F - kmAg -k(mA - mB)g] / mB= [20N - 0.4(1kg)(9.8m/s/s) - 0.4(1kg + 2kg)(9.8m/s/s)] / 2kg = 2.16 m/s2 I hope this helps!

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