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An 7.0 kg sled is initially at rest on a horizontal road. The coefficient of kinetic friction between the sled and the road is 0.40. The sled is pulled a distance of 3.0 m by a force of 45 N applied to the sled at an angle of 30� to the horizontal.

(a) Find the work done by the applied force.

(b) Find the energy dissipated by friction.

(c) Find the change in the kinetic energy of the sled.

(d) Find the speed of the sled after it has traveled 3 m.

Added Mon, 03 Aug '15

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total displacement = 3 m

a) work done by applied force = F*d cos theta = 45*3*cos 30 = 116.913 J

b) energy disspation by friction = - work done by friction = f d

friction = mu* ( mg- F sin theta ) = 0.4*( 7*9.81- 45 sin 30) = 18.468

so ,

work done by friction = - 3* 18.468 = -55.404 J

so , energy disspation by friction= 55.404 J

c) change in kinetic energy = sum of work done by all forces (work energy theorem)

change in KE =W (force) + W(friction ) = 116.913 - 55.404 =61.509 J

d)initial KE = 0., therefore ,

final KE = change in KE = 61.509 J,

1/2 m* v^2 = 61.509,

v^2 = 61.509*2/7,

final velocity , v = 4.192 m/s

Added Mon, 03 Aug '15

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