A package of mass m = 0.90kg is pushed across the surface by a jet of compressed

Question

A package of mass m = 0.90kg 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 = 49.0cm . The package and the surface are known to have a coefficient of kinetic friction of ?k = 0.10 . The package is initially at rest for this process.

Part A - Determining the average force supplied by the jet of compressed air If the package must be moving at a speed of v = 15.0 cm/s when it has reached the distance of 49.0 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 49.0 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 = 10.5 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 in = 1.35 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 mu k = 0.19. 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)

We will use conservation of energy to solve this question,

Let the average force of the jet be F

So, work done by force, WF = F*d <--- here d = distance traveled = 49 cm = 0.49 m

work done by friction , Wf = -umgd <------NOTE : the -ve sign due to opposite direction for friction to direction of motion

where u = coefficient of friction = 0.1

m = 0.9

g = 9.8 m/s2

d = 0.49 m

So, net work done , Wnet = WF + Wf = Fd - umgd

By conservation of energy net work done must be equal to change of energy,

So, Wnet = 0.5*m*(Vf^2-Vi^2) = Fd - umgd

So, 0.5*0.9*(0.15^2-0^2) = F*0.49 - 0.1*0.9*9.8*0.49

So, F = 0.903 N <----------answer

B)

for Vi = 10.5 cm/s = 0.105 m/s

Wnet = 0.5*m*(Vf^2-Vi^2) = Fd - umgd

So, 0.5*0.9*(0.15^2 - 0.105^2) = F*0.49 - 0.1*0.9*9.8*0.49

So, F = 0.893 N <-------------answer

C)

Now , m = 1.35 kg

u = 0.19

We know,

Wnet = 0.5*m*(Vf^2-Vi^2) = Fd - umgd

So, 0.5*1.35*(0.15^2 - 0.105^2) = F*0.49 - 0.19*1.35*9.8*0.49

So, F = 2.53 N <------------answer

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