A skier of mass 82.9 kg starts from rest at the top of a frictionless incline of height 20 m. At the bottom of the incline, the
1 answer:
Answer: 170.67 N
Explanation:
Given
Mass of skier is
Height of the inclination is
Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.
Thus, the horizontal friction force is 170.67 N.
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Answer:
(a) -472.305 J
(b) 1 m
Explanation:
(a)
Change in mechanical energy equals change in kinetic energy
Kinetic energy is given by
Initial kinetic energy is
Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero
Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence
0-472.305 J=-472.305 J
(b)
From fundamental kinematic equation
Where v and u are final and initial velocities respectively, a is acceleration, s is distance
Making s the subject we obtain
but a=\mu g hence
Answer:
The speed of the student just before she lands, v₂ is approximately 8.225 m/s
Explanation:
The given parameters are;
The mass of the physic student, m = 43.0 kg
The height at which the student is standing, h = 12.0 m
The radius of the wheel, r = 0.300 m
The moment of inertia of the wheel, I = 9.60 kg·m²
The initial potential energy of the female student, P.E.₁ = m·g·h₁
Where;
m = 43.0 kg
g = The acceleration due to gravity ≈ 9.81 m/s²
h = 12.0 h
∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J
The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;
The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0
∴ K₁ = 0 + 0 = 0
The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0
Where;
h₂ = 0 m
The final kinetic energy, K₂, of the wheel and student is give as follows;
Where;
v₂ = The speed of the student just before she lands
ω₂ = The angular velocity of the wheel just before she lands
By the conservation of energy, we have;
P.E.₁ + K₁ = P.E.₂ + K₂
∴ m·g·h₁ + = m·g·h₂ +
Where;
ω₂ = v₂/r
∴ 5061.96 J + 0 = 0 +
5,061.96 J = 21.5 kg × v₂² + 53. kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²
v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²
v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s
The speed of the student just before she lands, v₂ ≈ 8.225 m/s.