a 96 kg clock initially at rest on a horizontal floor requires a 605 N horizontal force to set it in motion. After the clock is in motion a horizontal force of 547 N keeps it moving with a constant velocity. the acceleration of gravity is 9.81 m/s2 find the static friction between clock and floor

Answer 1

Wc = mg = 96kg * 9.8N/kg = 940.8 n. =
Wt. of clock.

Fc = 940.8 N. @ 0o = Force of clock.
Fp = 940.8*sin(0) = 0 = Force parallel
to floor.

Fs = Force of static friction.

Fap-Fp-Fs = ma.
605-0-Fs = 96*0 = 0.
-Fs = -605
Fs = 605 N.

Answer 2

891.9