A 5.35 kg package slides 1.25 m down a long ramp that is inclined at 13.0 degree

Question

A 5.35 kg package slides 1.25 m down a long ramp that is inclined at 13.0 degree below the horizontal. The coefficient of kinetic friction between the package and the ramp is mu_k= 0.310. Calculate the work done on the package by friction. Calculate the work done on the package by gravity. Calculate the work done on the package by the normal force. Calculate the total work done on the package. If the package has a speed of 2.6 m/s at the top of the ramp, what is its speed after sliding 1.25 m down the ramp?

Explanation / Answer

normal force N = m*g*costheta

friction f = uk*N = uk*m*g*costheta

part(a)

work done by friction Wf = fk*L*cos180 = -0.13*5.35*9.8*cos13*1.25 = -8.3 J

part(b)

Wgravity = m*g*sintheta*L = 5.35*9.8*sin13*1.25 = 14.74 J

part(c)

angle netween normal and distanc travelled = 90

WN = N*l8cos90 = 0

part(d)

Wtotal = -8.3+14.74 + 0 = 6.44 J

part(e)

from work energy theorem

wtotal = change in KE

Wtotal = (1/2)*m*(vf^2-vi^2)

6.44 = (1/2)*5.35*(vf^2-2.6^2)

vf = 3.03 m/s

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