An object, which is initially at rest on a frictionless horizontal surface, is a

Question

An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces. F1 is 21.6 N acting due East, F2 is 41.2 N acting due North, F3 is 39.2 N acting due West, and F4 is 14.1 N acting due South. How much total work is done on the object in 2.22 seconds, if it has a mass of 22.0 kg? type of energy is changing for the object while the above work is being done? elastic potential energy gravitational potential energy kinetic energy internal energy How fast does the object end up moving at the end of the 2.22 seconds?

Explanation / Answer

net force = Fnet = Fx-Fy

Fx =
21.6-39.2 = -17.6 N

Fy = 41.2-14.1= 27.1 N

Fnet = sqrt(Fx^2+Fy^2) = sqrt(17.6^2+27.1^2) = 32.2 N

acclearation a =Fnet /m = 32.3/22 = 1.46 m/s^2

then apply v = u+ (a*t)

initial speed is u = 0 m/s

a = 1.46 m/s^2

t = 2.22 s

then v = 0+(1.46*2.22) = 3.26 m/s

total work done is change in kinetic energy

W = 0.5*m*(v^2-u^2) = 0.5*22*3.26^2 = 117 J

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kinetic energy

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v = 3.26 m/s

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