To be able to use the conservation of energy on a moving body to determine key c

Question

To be able to use the conservation of energy on a moving body to determine key characteristics about the mechanics of the system.

A package of mass m = 0.70 kg is pushed across the surface by a jet of compressed air during an assembly-line process. The package is moved through a distance of s = 48.0 cm . The package and the surface are known to have a coefficient of kinetic friction of ?k = 0.12 . The package is initially at rest for this process.

(Figure 1)

Part A - Determining the average force supplied by the jet of compressed air

If the package must be moving at a speed of v = 20.0 cm/s when it has reached the distance of48.0 cm , determine the average force that the jet of compressed air must supply to the package

Express your answer to three significant figures and include the appropriate units.

Part B - Determining the average force supplied by the jet of compressed air for nonzero initial velocity

If it is now desired that the package be at rest when it has reached the distance of 48.0 cm , determine the average force that the jet of compressed air must supply to the package if the package is initially traveling in the s direction at a velocity of v = 20.0 cm/s .

Express your answer to three significant figures and include the appropriate units.

Part C - Determining the average force supplied by the jet of compressed air for a different package

If the plant has changed the mass of the package on the assembly line to m = 1.40 kg , determine the average force that the jet of compressed air must supply to the package if the package and surface now have a coefficient of kinetic friction of ?k = 0.18 . Use the same initial and final conditions as in Part B.

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

A) friction force = u N = u mg = 0.12 x 0.7 x 9.8 = 0.8232 N

Applying work energy theorem,

Work done by Force + work done by friction = change in KE

( F (0.48)) + ( - 0.8232 x 0.48 ) = 0.70 x 0.20^2 /2 - 0

0.48F = 0.409

F = 0.852 N

B) Applying work energy theorem,

Work done by Force + work done by friction = change in KE

( F (0.48)) + ( - 0.8232 x 0.48 ) = 0 - 0.70 x 0.20^2 /2

0.48F = 0.3811

F = 0.794 N

c) Now mass and friction coefficient has changed.

friction force = u mg = 0.18 x 1.40 x 9.8 = 2.4696 N

Applying work energy theorem,

Work done by Force + work done by friction = change in KE

( F (0.48)) + ( - 2.4696 x 0.48 ) = 0 - 0.70 x 0.20^2 /2

F =2.44 N

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