10 POINTS: A steel cable drags a 22kg across a horizontal, frictionless surface.?

a = (v - vo)/t = (4.50 - 0)/2.1 = 2.143 m/s^2

Let T1 = tension at the end where the block is tied

T2 = tension at the end where the force is applied

T1 = m a = 22 x 2.143 = 47.146 N

T2 = 120 N

T2 - T1 = 120 - 47.146 = 73 N (rounded off)

Ans: 73 N

_________________.

First you have to find the mass of the cable.

Use m*a=F

(22kg+x)*(4.50/2.1)=120N

x= mass of cable

Now find the tension in the cable at 2 places: 1) at the end of the cable (where you would be holding it if you were pulling) and 2) where it is attached to the block.

1. T1=(mass of block)*a = 22*(4.50/2.1)

2. T2= (mass of block + mass of cable)*a = (22+x)(4.50/2.1)= 120N

T2-T1= difference in tension between 2 ends of cable.

if the exterior is frictionless there is not any stress on the cable as quickly as the block starts to bypass, you do no longer want a "massive" metallic cable to pull a 36lb block everywhere a single twine will do this , who thinks up those questions?