Work done against friction

Examples of friction:  rock climbing (static friction); baseball player sliding (sliding friction); sky diving (air friction); swimming (viscous friction); and car tire (rolling friction). Friction often converts kinetic energy into forms that are not easily converted back into motion. Kinetic friction on a skidding car or a sliding baseball player acts to oppose the motion, transforming kinetic energy into unrecoverable heat. A skydiver acted on by air resistance and a swimmer subject to drag are also examples of systems that do work on their surroundings through the action of a frictional force, losing mechanical energy in the process. Read the text aloud Show Friction and drag
In problems involving friction, work done against friction is energy that cannot appear in other forms. When you set up the conservation of energy equation, the work done against friction appears in the final state, after the change has occurred. Read the text aloud Show Where does the energy go?
What is speed of the hockey puck after traveling 20 m?
For example, a 170 g hockey puck with an initial speed of 10 m/s slides with a coefficient of friction of μk = 0.10. What is its speed after moving 20 m?
Asked: final speed v
Given: coefficient of kinetic friction μk = 0.10, mass of puck m = 0.17 kg,
initial velocity v0 = 10 m/s, distance traveled d = 20 m
Relationships: Ff = μkmg, W = Fd, Ek = ½mv2
Solution: 1 2 m v 0 2 = 1 2 m v 2 + μ κ mgd v 2 = v 0 2 2 μ k gd v= v 0 2 2 μ k gd v= ( 10 m/s ) 2 2( 0.10 )( 9.8  m/s 2 )( 20 m ) =7.8 m/s
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What is the change in the swimmer's kinetic energy? In many circumstances work is done against friction but the net change in kinetic energy is zero! For example, a swimmer expends 2,500 J of energy swimming 25 m across a pool. Friction exerts a force of 100 N against him.
  1. How much work is spent overcoming friction?
  2. What is the change in the swimmer’s kinetic energy?
Asked: (a) the work W done against friction and
(b) the change in kinetic energy ΔE
Given: ΔE = 2,500 J, Ff = 100 N, d = 25 m
Relationships: W = Fd, ΔE = Qfriction + W
Solution: (a) The work done against friction is W = Fd = (100 N)(25 m) = 2,500 J.
(b) The change in energy is ΔE = Q + W = 2,500 J − 2,500 J = 0.
There is no change in kinetic energy since all the work done by the swimmer (2,500 J) is spent overcoming friction.
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Katie pushes on a 10 kg box with 70 N of force, but the box only accelerates by 3 m/s2. How much friction is there?
  1. 0.3 N
  2. 7 N
  3. 20 N
  4. 40 N
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