A 1.50m long, 550g rope pulls a 7.00kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.20m/s2 . How much force pulls forward on (a) the ice, (b) the rope?
3 Answers

Rope Pulls

The ice accelerates @ 2.2m/s^2. It has a mass of 7 kg. F = ma
F = 7 kg x 2.2 m/s^2 = 15.4 N.
The ice / rope “system” has a mass of 7.55 kg. It is also accelerating at a rate of 2.2 m/s^2.
F = 7.55 kg x 2.2 m/s^2 = 16.61 N

solution;
mass of rope m=550g=0.55
mass of block mb =7 kg
acceleration produced by
block a=2.2m/s^2
now according to the formula
f=ma
=7*2.2
=15.4N
and, acceleration produced by rope means acceleration produced by block;
f=ma =[0.55+7]*2.2=16.61N
HENCE 15.4N force pulls forward the ice and 16.61N force to the rope
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