- #1

SakuRERE

- 68

- 5

## Homework Statement

A system, consisting of a wide rope of mass 0.10 kg between two blocks each of mass 0.10 kg, is lifted by an applied force F = 9.0 N (Fig. 4 below). (a) Find the acceleration of the system. Find the tension at (b) the top of the rope, and (c) the bottom of one-fifth of the rope. Take g = 10 m/s2.

## Homework Equations

Fnet=ma

## The Attempt at a Solution

a) Fnet=ma

F- 3mg = ma

9- 3(0.1)(10) = 0.1a

6=0.1 a

a= 60 m/s^2

b) T- 2mg = ma

T-2(0.10)(10) = 0.10(60) (it's right to use the mass of rope by acceleration right?)

T-2=66

Ttop= 68N

is it right?

c) T- (4/5x m)- 10xm= (4/5)m x a

T- 0.08(10) - (0.10)(10)= 0.08(60) is this equation right

i don't if i got (the bottom of one-fifth of the rope) correctly or not. i considered the tension to equal to that part of the rope below 1/5 from the mass so that m of rope now is 4/5m

and also i have the lower mass (m). so is it right??