Answer:
Ek = Ekv + Ekh = 4.101 + 0.914 = 5.015J
Explanation:
A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its potential energy (PE) when it lands
The potential energy PE, relative to the ground, will be zero, because the lemming is at the ground level.
HOWEVER, a much better question would be:
A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its kinetic energy (KE) when it lands?
Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):
s = ut + ½at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, u = 0, a = 9.81m/s^2, s = 5.36m
So we find v using equation (2)
v^2 = u^2 + 2as
v^2 = 0 + 2(9.81)(5.36) = 105.1632
So the kinetic energy resulting from the vertical drop is Ekv = ½mv^2
Ekv = ½(0.078)(105.16) = 4.101J
BUT we need to add in the kinetic energy resulting from the horizontal velocity, which did not change during the vertical drop.
Ekh = ½(0.078)(4.84^2) = 0.914J
So the total kinetic energy is Ek = Ekv + Ekh = 4.101 + 0.914 = 5.015J
