An animal climbs a fallen tree that is angled at 45 degrees with repsect to the

Question

An animal climbs a fallen tree that is angled at 45 degrees with repsect to the ground. If it climbs a distance along the tree trunk of 30 meters in 4.5 minutes , how much potential energy did she gain or lose by the end of the climb? What mechanical work did she perform during hte climb? The animal has a mass of 3.0 kg. If it slipped off the tree after her climb, how fast would she be going just before she landed on a thin soft pillow on the ground? Neglect Air Resistance?

Please help me also by explaining your thought process! Thanks!

Explanation / Answer

2*(side of triangle)^2 = 30^2 = 900

side = 21.2 m

so pot energy = mgh = 3*9.8*21.2 = 623.28 J

If we consider slow climb (which we must as the climbing action is not under a conservative force) , the work done = change in kinetic energy = 0

if she falls, change in kinetic energy = pot. energy

0.5*m*v^2 = 623.28

v = 20.38 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects