A package of mass 5.50 kg slides a distance 1.48 m down a long ramp that is incl

Question

A package of mass 5.50 kg slides a distance 1.48 m down a long ramp that is inclined at an angle 11.8 degrees below the horizontal. The coefficient of kinetic friction between the package and the ramp is mu_k = 0.305.

Part A
Calculate the work done on the package by friction.

Part B
Calculate the work done on the package by gravity.

Part C
Calculate the work done on the package by the normal force.

Part D
Calculate the total work done on the package.

Part E
If the package has a speed of 2.26 m/s at the top of the ramp, what is its speed after sliding the distance 1.48 m down the ramp?

Explanation / Answer

mass m = 5.50 kg distance S = 1.48 m angle theta = 11.8 degrees The coefficient of kinetic friction between the package and the ramp is = U= 0.305 (A).the work done on the package by friction = -Umg cos(90-theta) * S W1= -4.97 J ( B).the work done on the package by gravity = mg sin(90-theta) * S W2 = 78.08 J ( C).the work done on the package by the normal force W3 = 0 Since normal force and displacment are perpendicular to each other. ( D).the total work done on the package W = W1+ W2 + W 3 W = 73.11 J ( E).Initial speed u = 2.26 m/s speed after sliding the distance 1.48 m down the ramp v = ? from work energy theorem , W = ( 1/ 2) m[ v^ 2- u^ 2] from this v ^2= (2W / m) + u^ 2 = 31.69 v = 5.62 m / s

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