A 5.20 kg package slides 1.52 m down a long ramp that is inclined at 12.0 below

Question

A 5.20 kg package slides 1.52 m down a long ramp that is inclined at 12.0 below the horizontal. The coefficient of kinetic friction between the package and the ramp is k = 0.310

Part A

Calculate the work done on the package by friction.

Express your answer to three significant figures.

Part B

Calculate the work done on the package by gravity.

Express your answer to three significant figures.

Part C

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

Express your answer to three significant figures.

Part D

Calculate the total work done on the package.

Express your answer to three significant figures.

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 1.52 m down the ramp?

Express your answer to three significant figures.

Explanation / Answer

A. friction force, f = uk N = uk m g cos@

f = 0.310 x 5.20 x 9.8 x cos12 = 15.45 N

Work done = F.d = Fdcos@

where @ is the angle between F and d.

@ = 180 deg

W = 15.45 x 1.52 x cos180 = - 23.5 J

B. W = mg d cos(90-12) = 5.20 x 9.8 x 1.52 x sin12 = 16.10 J

C. for normal force , @ = 90

cos90 = 0

W = 0

D. Wnet = -23.5 + 16.10 + 0 = - 7.4 J

E. total work done = change in KE

- 7.4 = 5.20 ( v^2 - 2.26^2) / 2

- 2.84 = v^2 - 2.26^2

v = 1.506 m/s

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