Where Did I Go Wrong With the Rope and Ice Force Calculation?

[mullets1200]

Homework Statement

A 1.50 m-long, 480 g rope pulls a 8.00 kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.40 m/s^2. How much force pulls forward on (a) the ice, (b) the rope?

Homework Equations

F=m*a

The Attempt at a Solution

for A i got the right answer by
Acceleration of the block = 2.4 m/s^2
mass of the block = 8kg
Force = 8*2.4 = 19.2 N

but for b i did:
Acceleration of the rope = Acceleration of the block = 2.4 m/s^2
mass of the rope = 0.480kg
Force = 0.480*2.4 = 1.152 N
and it was wrong. anyone know where i messed up?

 

[Delphi51]

The pull on the rope must accelerate not only the rope but the ice as well.
Add the 1.152 onto the 19.2

Another way to look at it is that the rope has unknown F pulling it forward, and 19.2 N pulling backward. Using sum of forces = ma, this is
ma = F - 19.2
F = ma + 19.2

 

[rl.bhat]

Force on the rope = Applied force - reaction force by the block.

 

[mullets1200]

oh! i don't know why i didn't catch that. thank you so much!