A 6.00-kg package slides 2.80 m down a long ramp that is inclined at 23.0 below

Question

A 6.00-kg package slides 2.80 m down a long ramp that is inclined at 23.0 below the horizontal. The coefficient of kinetic friction between the package and the ramp is k = 0.320.

Part A:

Calculate the work done on the package by friction.

Express your answer with the appropriate units

Part B:

Calculate the work done on the package by gravity.

Express your answer with the appropriate units.

PART C) Calculate the work done on the package by the normal force.

PART D)

Calculate the total work done on the package.

If the package has a speed of 2.50 m/s at the top of the ramp, what is its speed after it has slid 2.80 m down the ramp?

Explanation / Answer

a) Work done by friction = uk*mg*cos23* 2.8m = 48.49 J

b) Work done by gravity = mg*sin23* 2.8 m= 64.33 J

c) zero .. Angle between displacement and normal force is 90 so cos 90 is zero

d) Change in KE = work done

1/2 m ( vf^2 - 2.5^2) = 64.33-48.49

vf = 3.3 m/s

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