don\'t know what to do An object, which is initially at rest on a frictionless h
ID: 2142598 • Letter: D
Question
don't know what to do
An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces. F1 is 28.6 N acting due East, F2 is 41.2 N acting due North, F3 is 43.5 N acting due West, and F4 is 10.8 N acting due South. How much total work is done on the object in 2.22 seconds, if it has a mass of 12.0 kg? Which type of energy is changing for the object while the above work is being done? How fast does the object end up moving at the end of the 2.22 seconds?Explanation / Answer
We do the force balance:
F1 and F3 oppose each other
F2 and F4 oppose each other
Thus net forces =
F3 - F1 = 43.5 - 28.6 = 14.9 N due west = Fw (say)
F2 - F4 = 41.2 - 10.8 = 30.4 N due North = Fn (say)
Thus net force acting (magnitude) = sqrt (Fw^2 + Fn^2) = 33.85 N
F = M * a {Newtons law}
Thus, acceleration of the body = F / M = 33.85 / 12.0 = 2.82 m/s^2
Since acceleration is a constant, we can use equations of kinematics
Thus, velocity v = u + at
u =initial velocity, v= final velocity, a = acceleration, t= time
u = 0, {initially at rest}
Thus v = at = 2.82 (2.22) = 6.26 m/s is the velocity at the end of 2.22 sec
Displacement = s = ut + 0.5at^2 = 0 + 0.5(2.82)(2.22)^2 = 6.95 m at the end of 2.22 sec
Work done = Force * displacement = 33.85 * 6.95 = 235.30 Joules of work is done.
This work is obtained by CHANGE IN KINETIC ENERGY OF THE BODY.
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