Bored, a boy shoots his pellet gun at a piece of cheese that sits, keeping cool for dinner guests, on a massive block of ice. on one particular shot, his 1.3 g pellet gets stuck in the cheese, causing it to slide 25 cm before coming to a stop. if the muzzle velocity of the gun is 73 m/s and the cheese has a mass of 109 g, what is the coefficient of friction between the cheese and ice?

Answer

mass of pellet (m₁) =1.3 g

mass of the system when pellet sticks to the cheese is (m₂)

m₂ = 109 + 1.3 = 110.3 g

velocity of the muzzle = 73 m/s

now using conservation of momentum  

m₁ v₁ =m₂v₂

where v₁ is velocity of the pellet and v₂ is the initial velocity of the (pellet+cheese ) system

 m₁ v₁ =m₂v₂

1.3 x 73 = 110.3 x v₂

v₂ = 0.86 m/s

initial velocity of (pellet+cheese)system = v₁

and final velocity =  0

using

v₁² - 0 = 2 a s

where a is the deceleration and s =25 cm

0.86² - 0 = 2 a s

0.86² - 0 = 2 x a x 0.25

a = 1.48 m/s²

now, calculation of force of friction

m₂ a = μ m₂ g

where  μ is the coefficient of friction

μ = 0.087

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever