Question

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?

Answer 1

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Step-by-step explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

`v=u+at\\\Rightarrow 0=5-9.81* t\\\Rightarrow (-5)/(-9.81)=t\\\Rightarrow t=0.51 \s`

`s=ut+(1)/(2)at^2\\\Rightarrow s=5* 0.51+(1)/(2)* -9.81* 0.51^2\\\Rightarrow s=1.27\ m`

So, the stone would travel 1.27 m up

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* 9.81* 1.27+0^2)\\\Rightarrow v=5\ m/s`

Velocity as the rock passes through the original point is 5 m/s

`s=ut+(1)/(2)at^2\\\Rightarrow 1.27=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(1.27* 2)/(9.81)}\\\Rightarrow t=0.51\ s`

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

`s=ut+(1)/(2)at^2\\\Rightarrow 16.27=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(16.27* 2)/(9.81)}\\\Rightarrow t=1.82\ s`

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

`v=u+at\\\Rightarrow v=0+9.81* 1.82\\\Rightarrow v=17.8542\ m/s`

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

`s=ut+(1)/(2)at^2\\\Rightarrow 31.27=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(31.27* 2)/(9.81)}\\\Rightarrow t=2.52\ s`

`v=u+at\\\Rightarrow v=0+9.81* 2.52\\\Rightarrow v=24.7212\ m/s`

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

`745.7=(\mu * 981* 4000)/(1200)\\\Rightarrow \mu=(747.5* 1200)/(981* 4000)\\\Rightarrow \mu=0.229`

Coefficient of sliding friction is 0.229