component of weight normal to plane
Fn = m g cos 11.6 = 51.9 N
component of weight down the ramp
Fd = m g sin 11.6 = 10.7 N
Ff = Friction force = .313*Fn = 16.2 N
work done by friction = -1.47 Ff = -23.9 Joules (force opposite to motion so negative)
work done by gravity = m g * height change = m g * 1.47 sin 11.6
= 15.7 Joules
Normal force is perpendicular to motion so does no force
Total work
= 15.7 - 23.9 = - 8.2 Joules
Ke at end = Ke at start -8.2 Joule
(1/2) m v^2 = (1/2) m(2.15)^2 -8.2
.5 m v^2 = 4.28 Joules
v = 1.26 m/s
A 5.40 Kg package slides 1.47 meters down a long ramp that is inclined at 11.6 degress below the horizontal. The coefficient of kinetic friction between the package and the ramp is = 0.313. 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 2.15 m/s at the top of the ramp, what is its speed after sliding the distance 1.47 meters down the ramp?
1 answer
1 answer