A 20 kg box is being pulled across a floor by a horizontal rope. The tension in the rope is 99 Newtons. The coefficient of frict
2 answers:
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Answer:
48.16 %
Explanation:
coefficient of restitution = 0.72
let the incoming speed be = u
let the outgoing speed be = v
kinetic energy = 0.5 x mass x
- incoming kinetic energy = 0.5 x m x
- coefficient of restitution =
0.72 =
v = 0.72u
therefore the outgoing kinetic energy = 0.5 x m x
outgoing kinetic energy = 0.5 x m x
outgoing kinetic energy = 0.5184 (0.5 x m x )
recall that 0.5 x m x is our incoming kinetic energy, therefore
outgoing kinetic energy = 0.5184 x (incoming kinetic energy)
from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.
The energy lost would be 100 - 51.84 = 48.16 %
Hi there!
We can use the kinematic equation:
vf = Final velocity (? m/s)
vi = initial velocity (0 m/s, dropped from rest)
a = acceleration (due to gravity, 9.8 m/s²)
d = distance (9.8 m)
Simplify the equation to solve for vf:
Substitute in the given values: