The problem is that there is a 2 kg block with a 1kg block ontop of it. The 2 kg
ID: 1667217 • Letter: T
Question
The problem is that there is a 2 kg block with a 1kg block ontop of it. The 2 kg block is being pulled to the right with atension force of 20 N and the 1kg block is connected to a rope thatis connected to the wall to the left. The coefficients of kineticfriction are 0.40 between the ground and 2kg block and 2kg blockand 1kg block. The questions are a) what is the tension in the rope connected to the 1kgblock b) what is the acceleration of the 2kg blockI tried setting it up by using Newtons 2nd law. I get Fnet(1kg) = 0 = -T + Ffriction Fnet(2kg) = T(20N) - Ffriction(ground) - Ffriction(1kg) =ma
I dont know how to figure out the tension in the block. Imight have my directions for friction wrong too. And for Normalforce for the bottom block, do i have to use 3kg, since its totalmass is both blocks combined? The problem is that there is a 2 kg block with a 1kg block ontop of it. The 2 kg block is being pulled to the right with atension force of 20 N and the 1kg block is connected to a rope thatis connected to the wall to the left. The coefficients of kineticfriction are 0.40 between the ground and 2kg block and 2kg blockand 1kg block. The questions are a) what is the tension in the rope connected to the 1kgblock b) what is the acceleration of the 2kg block
I tried setting it up by using Newtons 2nd law. I get Fnet(1kg) = 0 = -T + Ffriction Fnet(2kg) = T(20N) - Ffriction(ground) - Ffriction(1kg) =ma
I dont know how to figure out the tension in the block. Imight have my directions for friction wrong too. And for Normalforce for the bottom block, do i have to use 3kg, since its totalmass is both blocks combined?
Explanation / Answer
Your second equation is right as well and for normal force youneed to take into account the mass of both blocks
20N - (3kg*9.8*0.4) - (1kg*9.8*0.4) = 2kg * a
4.32N/2kg = 2.16N/kg
Hope that helps
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