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A 1.50m -long, 550g rope pulls a 7.00kg block of ice across a horizontal, frictionless surface. Physics Help?

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

3 Answers

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

A
Anonymous

Feb 05, 2025

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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