The figure shows two 1.0 kg blocks connected by a rope. a second rope hangs beneath the lower block. both ropes have a mass of 250

g. the entire assembly is accelerated upward at 3.0 m/s2 by force fâ -.

We need first to use the formula F=m (a+g), m iis the total mass, a is the acceleration, g is gravity pulling the blocks. So the procedure will be

m=2kg (both blocks) + 500g (both ropes) → m=2.5kg

a=3.00m/s^2

g=9.8m/s^2

F=m (a+g) → F=2.5kg (3.00m/s^2 + 9.8m/s^2) → F=2.5kg (12.8m/s^2) → F=32 N

To calculate the tension at the top of rope 1 you need to use the formula T=m (a+g) so it will be T=m (a+g) → T=1.5kg (12.8m/s^2) → T=19.2N

We can now calculate the tension at the bottom of rope 1 using the formula: T=m (a+g) → T=1.25kg (12.8m/s^2) → T=16N

Now to find the tension at the top of rope 2 we do it like this:

T=m (a+g) → T=.25kg (12.8m/s^2) → T=3.2