I know this is simple but Im having problems A package of mass 5.50 kg slides a

Question

I know this is simple but Im having problems
A package of mass 5.50 kg slides a distance 1.57 m down a long ramp that is inclined at an angle 11.0^circ below the horizontal. The coefficient of kinetic friction between the package and the ramp is mu_k = 0.300.
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 2.26 m/s at the top of the ramp, what is its speed after sliding the distance 1.57 m down the ramp?

Explanation / Answer

The mass of package, m = 5.5 kg Distance moved, s = 1.57 m angle of inclination, = 11 degrees coefficient of kinetic friction, k = 0.3 Workdone by the friction = fs = mgcos * s = 24.9 J Workdone by the gravity = fs = mgsin * s = 16.15 J Workdone by normal force = always zero Total workdone = 24.9 + 16.15 = 41.05 J The initial velocity, u = 2.26 m/s Final velocity = v distance, s = 1.54 m The slides down on inclined plane, so acceleration, a = g( sin - k cos )                             a = 9.8( sin11 - 0.3cos11 ) = - 1.02 m/s^2          We have, v^2 = 2as                          v = sqrt(2as) = sqrt(2*1.02*1.54) = 1.77 m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---