A box of mass 16.6 kg with an initial velocity of 1.48 m/s slides down a plane,

Question

A box of mass 16.6 kg with an initial velocity of 1.48 m/s slides down a plane, inclined at 33 degrees with respect to the horizontal. The coefficient of kinetic friction is .96. The box stops after sliding a distance x. The acceleration due to gravity is 9.8 m/s^2. The positive x direction is down the plane. a) What is the magnitude of the instantaneous power generated by friction half way between the initial and final positions? Answer in units of W. b) What is the magnitude of the average power generated by friction from start to stop? Answer in units of W.

Explanation / Answer

a)friction force = *N= *mgcos33 =130.977 N

net acceleration a_net= gsin33-gcos33 = -2.553 m/s^2

t_stop=u/a_net =0.5797 s (v=u+at)

x=0^2-u^2/(2*a_net) = 0.42898 m (v^2-u^2=2as)

when s=x/2 v^2-1.48^2=2*(-2.553)*(0.42898/2) =

v=1.0952 m/s

a)instantaneous friction power= friction force*velocity =143.447 W

b)average friction power= total energy dissipated /time taken =(130.977*0.42898)/0.5797 = 96.9234 W

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