Physics work being done?

A) W=Fdcosθ, force done by the rope is equivalent to the force of friction so the magnitude of the horizontal forces can be zero (balanced) and speed can remain constant (Newton's First Law). Force of friction = F(normal)*(coefficient) = mg*(coefficient) = 50kg*(9.8m/s^2)*(.15) = 73.5N. W=73.5N(150m)cos(40)

=8446Nm

B) W=Fdcosθ, the normal force vector is 90 degrees to the displacement vector. cos(90) = 0. W=Fd(0)

= 0Nm

C) The normal force is the equal and opposite force that the Earth exerts back on the object (Newton's Third Law). Since the object is pushing straight down on the earth (not on a slope), the normal force = force of gravity. So answer is the same as B)

= 0Nm

D) W=Fdcosθ, we calculated force of friction in A). W=73.5N(150m)cos(180) = 73.5N(150m)(-1)

= -11025Nm

E) Total work done is simply the magnitude or sum of all the work done on the system. so ΣW=W(rope) + W(friction) = 8446Nm + (-11025Nm) = -2579Nm

= -2579Nm

captivating? Hm, that's no longer the first situation that contains ideas, besides the actual undeniable reality that something in nature may have a cultured personality. Awe inspiring easily. I particularly see quantum mechanics as particularly terrifying and punctiliously alien, a window right into a international so unusual that our brains are basically no longer in a position to perfect visualise it. certainly it really is unvisualiseable because it takes position at scales shorter than the wavelengths of light. possibly there is attractiveness in a sea of frothing microscopic chaos interior a similar way there is attractiveness in a backyard left to advance loopy and wild, notwithstanding that's not a classical attractiveness.

A. Work done by rope = work done by friction + 1/2* m* v^2

B. 0

C. 0

D. 0.15* m* g* 150

E. Total work done = work done by the rope