Work by conservative force (force of gravity is
W = m•g•h = 25•9.8•15 = = 3675 J.
KE = 1700 J.
W(fr) = W - KE = 3675 - 1700 = 1975 J.
Starting from rest at a height equal to the radius of the circular track, a block of mass 25 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient μ). The radius of the track is 15 m.
The acceleration of gravity is 9.8 m/s2 .
Determine the work done by the conservative forces. Answer in units of J.
If the kinetic energy of the block at the bottom of the track is 1700 J, what is the work done against friction? Answer in units of J.
1 answer
1 answer