(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 100 kg trunk loaded with old books is to be slid across a floor by a woman who exerts a force of 500 N down and forward at 30° with the horizontal. If μk = 0.3 and μs = 0.4, compute the resulting acceleration.

2. Relevant equationsand3. The attempt at a solution

N = mg, of trunk

N = 100kg * 9.8 = 980 N

Push of lady :

N = F sin theta

N = 500N * sin 30 = 250N

Total force being exerted = 980 + 250N = 1230N

friction kinetic force:

f = u*N

f = .3 * 1230N

friction force of moving box = 369N

woman pushes with force:

500N * cos 30 = 433N

So with 433N force of push - 369N friction = 64N of net force on sliding box

Net force = mass * acceleration

64N = 100kg * a

acceleration = 0.64 m/s^2 => INCORRECT => Please help!

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# Homework Help: Compute the resulting acceleration

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