The mechanical energy isn't conserved. Some energy is lost to friction.
Option A.
<h3><u>Explanation:</u></h3>
The mechanical energy is defined as the energy of a body which it achieves by virtue of its position and velocity. The mechanical energy are of two types - potential energy and kinetic energy. The potential energy is the energy of the body which it achieves by means of its relative position and is directly proportional to the height of the body from its relative plane. Whereas the kinetic energy of the body is achieved by virtue of its velocity and is directly proportional to the square of velocity of the body.
As the mountaineer is skiing down the slope of a mountain, the potential energy of the person is gradually changing into his kinetic energy. Had it been in an ideal situation, the potential energy lost would have been just equal to the kinetic energy gained by the person. But there's friction which opposes the speed of the body and reduces the velocity. Thus the kinetic energy will be lost to some extent and the energy won't be conserved.
Answer:
8.75
Explanation:
First, find the force of friction.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.9 m/s)² = F (1.4 m)
F = 11.7 N
Next, find the distance at the new velocity.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d
d = 8.75 m
Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F = 
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
