A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back past its original point, (b) When it is 15 m above the street, and (c) Just before it hits the street. A horse drags a 100 kg sled a distance of 4 km in 20 minutes. The horse exerts one horsepower, of course. What is the coefficient of sliding friction between the sled and the ground?

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s[/tex]

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s[/tex]

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

[tex]v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s[/tex]

Speed of the ball at 15 m above the ground is 17.8542 m/s

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s[/tex]

[tex]v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s[/tex]

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

[tex]745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229[/tex]

Coefficient of sliding friction is 0.229