Question

An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces. F is 21.6 N acting due east, F2 is 28.6 N acting due north, F3 is 43.5 N acting due west, and Fl is 10.8 N acting due south. How much total work is done on the object in 2.22 s, if it has a mass of 22.0 kg? total work done on object: Which type of energy is changing for the object while the work is being done? internal energy O elastic potential energy O gravitational potential energy kinetic energy How fast does the object end up moving at the end of the 2.22 s? E final speed of object:

Answer 1

(North) 28-6N 18- 28.6N f = 43.50 me M = 22 kg (West) m F, -21.6 M (East) F = 10.8N (South) t = 2.22 sec So, Net face F - - - Net force F = 17.8N m F"21.89 f = √ F1² + F12 117.8)+(21.9) - 796.45 F = 28.22 N Lin north-west direction)

Now, acceleration (a) F. m 28. 22 22 : a = 1.282 mpg² displacement (S) = ut & fat² here; u=0 += 2. 22 sec. - ☺ Loi s=o+tx 1.2826 (2.22² 1$ = 8.1591 M & work done on object = F xs. = 29.22 x 3.1591 Tw = 89.149 J Calong north west)

6.) kinetic energy . using first equation of motion; & V=utat & U = 0+ 1.282x (2.22) from quo & ) 10 = 2.846 m/sec. |
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