Respuesta :
Answer:
(a) ΔE[tex]_{th}[/tex] = 153.05 J
(b) Her speed at the bottom is 5.465 m/s
Explanation:
Here, we have
Slope of incline = 20 °
Normal force to incline = mg cos θ = F cos θ = 267 × cos 20 = 250.9 N
Force of friction = Kinetic friction × Normal force = 0.1 × 250.9 N = 25.09 N
If all the work done by the child sliding down the inclined plane is converted into thermal energy, we have
Work done = 25.09 N × 6.10 m = 153.05 J
The heat energy transferred to thermal energy. ΔE[tex]_{th}[/tex] = 153.05 J
(b) Here we have the some of the energies at the bottom is given by
ΔU + ΔK.E + ΔE[tex]_{th}[/tex] = 0
Where:
ΔU = Gravitational potential energy
ΔK.E = Change in kinetic energy of the child
ΔE[tex]_{th}[/tex] = Change in thermal energy of the child
ΔU = m·g·h
Where:
m·g = Weight of the child = 267 N and
h = Height from which the child starts the slide = Slope height
Slope height = 6.1 × sin 20° = 2.09 m
∴ ΔU = m·g·h = 267 N × 2.09 m = 557.05 J
Therefore. from ΔU + ΔK.E + ΔE[tex]_{th}[/tex] = 0, we have
-557.05 J + ΔK.E + 153.05 J = 0
ΔK.E = 557.05 J - 153.05 J = 404.0 J
However, ΔK.E = (0.5×m×(v₂²-v₁²)
Where,
v₁ = Initial velocity of child = 0.457 m/s
v₂ = Final velocity of child
Where 267 = m×g
Where:
g = Acceleration due to gravity = 9.8 m/s²
m = Mass of child
m = 267/g = 267/9.8 = 27.245 kg
404.0 J = (0.5×27.245×(v₂²-0.457²)
29.66 = v₂²-0.457²
v₂² = 29.66 + 0.457² = 29.866 m²/s²
v₂ = 5.465 m/s
Her speed at the bottom = 5.465 m/s.
