A 10 kg crate is sliding at a constant speed of 13 m/s along a frictionless hori

Question

A 10 kg crate is sliding at a constant speed of 13 m/s along a frictionless horizontal table, when it encounters a friction patch ( Mu k=0.3). a.) Find the work done by each force acting on the crate, after it has slid 5 m: work done by normal force: __J work done by gravity: __J work done by friction: __J b.) Find the speed of the crate after sliding this distance. ___ m/s c.) Some time later, the crate comes to rest. How much total work is done on the crate, from start to finish?* (from when it enters the friction patch to when it stops) ___ J Mu s=0.7,

Explanation / Answer

a) work done by normal force = 0 ( as the displacemnt is in perpendicular direction)

work done by gravitational force = 0 ( as the displacemnt is in perpendicular direction)

W friction = frictio force * d = mu(k) * m * g * 5 = -0.3*10* 9.81* 5

W (friction)= -147.15 J

b)

ma = friction = - mu(k) * m * g ,

so ,

a= -0.3 * g = -2.943 m/s^2

so ,

using , v^2 = u^2+ 2as

v ( after 5 m ) = sqrt ( 13^2- 2.943* 2 * 5)

so ,

speed after 5 m ,

v =11.814 m/s ~ 11.8 m/s

c)total work done = change in KE( from work -energy theorem)

so ,
total work done = 1/2 * 10* ( 13^2)

total W = 845 J

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