what is the speed of the object as it reaches the bottom of the incline? 10 pts?

work-energy theorem:

=> Δ k.e = net work done = net force x distance moved

net force = Fᵣₑ = 16gsinθ° - 6 = 16(9.81)(7/13) - [6] = 78.517 N

=> ½mv² - ½m0² = 78.517 x 13 = 1020.72

=> v² = 127.59 = 11.296²

=> v ~= 11.3 m/s

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or

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energy dissipated as heat doing work against friction = force x distance = 6 x 13 = 78 J

initial energy = potential energy = mgh = 16(9.81)7 = 1098.72 J

energy converted to kinetic energy(½mv²) on ground impact = 1098.72 - 78 = 1020.72 J

=> ½16v² = 1020.72 => v = √127.59 = 11.3 m/s


hope this helps

Taking g = 9.81 m/s²

(Loss in P.E.) = (Work done) + (Gain in K.E.)
m g h = F s + (1/2) m v²
16 × 9.81 × 7 = 6 × 12 + (1/2) × 16 × v²
v = √[(16 × 9.81 × 7 - 6 × 12) / 8] m/s

Speed of the object as it reaches the bottom of the incline = 11.3 m/s