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**Calculating energy drop**
I need to address the drop in Portuguese clear energy calculated as 60.3 within a series of numbers. I should take careful note of the negative dissipative magnitude in these computations. My response will focus on being concise with relevant formulas while performing the calculations, including initial, rest, potential, and kinetic energy. I won't get bogged down in unnecessary wording; my aim is efficiency and clarity, staying within the word limit.
A perda de altura até o ponto mais baixo é
\Delta h=70-9{,}70=60{,}30\ \text{m}.
Como Sandro parte do repouso, a energia potencial gravitacional perdida transforma-se em energia cinética, descontando a energia dissipada pelo atrito e pelo arrasto do ar:
\frac{mv^2}{2}=mg\Delta h-10\,710.
Substituindo os valores:
\frac{70v^2}{2}=70\cdot10\cdot60{,}30-10\,710
=31\,500.
Logo,
v^2=900 \quad\Rightarrow\quad v=30\ \text{m/s}.
Convertendo para km/h:
v=30\cdot3{,}6=\boxed{108\ \text{km/h}}.
Resposta: D
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levou 4 s
Como pensou
**Calculating work and velocity**
I’ll start by calculating the potential energy with the formula mgh: 70 kg * 10 m/s² * 60.3 m = 42,210 J. Next, I’ll subtract 10,710 J from this, which leaves me with 31,500 J. Now, for the velocity, I’ll take the square root of 900, which gives me 30 m/s. I should provide a brief explanation of negative work as well. This solution needs to be concise and straightforward!
Pelo teorema da energia cinética, o trabalho da gravidade, somado ao trabalho negativo das forças dissipativas, determina a energia cinética final. Como Sandro parte do repouso:
\frac{mv^2}{2}=mg(h_i-h_f)-10\,710.
A descida vertical é de 70-9{,}70=60{,}30\ \text{m}. Substituindo os valores:
\frac{70v^2}{2}=70\cdot10\cdot60{,}30-10\,710
=42\,210-10\,710=31\,500.
Assim,
v^2=900 \quad\Rightarrow\quad v=30\ \text{m/s}.
Convertendo para km/h:
v=30\cdot3{,}6=\boxed{108\ \text{km/h}}.
Resposta: D
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levou 6 s
Como pensou
**Calculating values**
I'm computing 70 times 10 times 60.3, which gives me 42,210. Then, I’m subtracting 10,710 from that result, leaving me with 31,500. Now, I see that equals 35 times v². I need to solve for v by dividing both sides by 35, which leads me to v² being 900. Taking the square root, I find v to be 30 m/s. The final value is 108.
A perda de energia potencial gravitacional na descida é
\Delta E_p = mg(h_i-h_f)
=70\cdot 10\cdot(70-9{,}7)=42.210\ \text{J}.
Descontando os 10.710 J dissipados pelo atrito e pelo arrasto do ar, a energia cinética no ponto mais baixo é
E_c=42.210-10.710=31.500\ \text{J}.
Como E_c=mv^2/2, temos
31.500=\frac{70v^2}{2}
\quad\Rightarrow\quad v=30\ \text{m/s}.
Convertendo para km/h: 30\cdot 3{,}6=108\ \text{km/h}.
Resposta: D
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levou 5 s
Como pensou
**Clarifying energy calculations**
I’m working on a problem involving energy in Portuguese. The initial height is 70m, and the bottom height is 9.7m, so that’s a drop of 60.3m. The potential energy loss is calculated as mg delta h, which comes out to 42,210 J. The work done against friction and air drag is -10,710 J, so the kinetic energy is 31,500 J. When I solve for the speed, it turns out to be 30 m/s, or 108 km/h. I need to make sure everything is stated concisely and clearly.
Pelo teorema da energia mecânica, a variação da energia cinética é igual ao trabalho da gravidade somado ao trabalho das forças dissipativas.
A queda vertical foi 70,0-9,7=60,3\ \text{m}. Assim, partindo do repouso:
K = m g \Delta h - 10\,710
=70\cdot10\cdot60,3-10\,710
=31\,500\ \text{J}.
Como K=\frac12mv^2,
\frac12\cdot70\cdot v^2=31\,500
\quad\Rightarrow\quad v=30\ \text{m/s}.
Convertendo para km/h: 30\cdot3,6=108\ \text{km/h}.
Resposta: D
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levou 6 s
Como pensou
Vamos aplicar o teorema da energia cinética (trabalho-energia) entre o ponto de partida (repouso, a 70,0 m) e o ponto mais baixo da rampa (a 9,7 m).
Desnível: Δh = 70,0 − 9,7 = 60,3 m
Trabalho do peso:
W_peso = m·g·Δh = 70 × 10 × 60,3 = 42.210 J
Trabalho das forças dissipativas (atrito e arrasto):
W_dis = −10.710 J
Energia cinética no ponto mais baixo:
Ec = W_peso + W_dis = 42.210 − 10.710 = 31.500 J
Velocidade:
½·m·v² = 31.500
v² = (2 × 31.500)/70 = 900
v = 30 m/s
Convertendo para km/h:
v = 30 × 3,6 = 108 km/h
Observação: a alternativa (B) 30 é uma "armadilha" para quem esquece de converter para km/h, e a alternativa (E) 126 corresponde a ignorar as forças dissipativas (v = √(2·10·60,3) ≈ 34,7 m/s ≈ 125 km/h).
Resposta: D
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levou 7 s
Como pensou
I'm working out the height drop as 60.3 m, which gives 31,500 J of kinetic energy and a speed of 30 m/s, or 108 km/h, so the answer is D. Now I'm wondering whether the 70 m is measured from the ground or from the bottom point, and checking the figure.
I'm confirming the 70 m is measured from the ground: the text says the platform sits 70 m above the ground, and the figure's dashed line runs down to the ground, so the drop height holds.
Resposta: (D) 108 km/h
Vou usar o teorema da energia: a energia potencial perdida na descida, menos o trabalho das forças dissipativas, vira energia cinética.
-
Desnível percorrido: Sandro parte de 70 m e chega ao ponto mais baixo a 9,70 m do solo.
Δh = 70 − 9,70 = 60,3 m
-
Energia potencial perdida:
m·g·Δh = 70 · 10 · 60,3 = 42.210 J
-
Desconto das perdas por atrito e arrasto:
Energia cinética no ponto mais baixo = 42.210 − 10.710 = 31.500 J
-
Velocidade (partindo do repouso):
(1/2) · 70 · v² = 31.500
35 · v² = 31.500
v² = 900
v = 30 m/s
-
Conversão para km/h:
30 m/s × 3,6 = 108 km/h
Resposta: D
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levou 5 s
Como pensou
I'm working out the drop height: 70 m from the platform to the ground, with the lowest point at 9.7 m, so the fall is 60.3 m. Using mgh gives 42210 J, and subtracting 10710 J leaves 31500 J of kinetic energy, which works out to a speed of 30 m/s.
The speed works out to 108 km/h, which matches option D.
Resolução
O skatista parte do repouso na plataforma, a 70 m do solo, e chega ao ponto mais baixo da rampa, a 9,70 m do solo. A altura percorrida na descida é, portanto:
Δh = 70 − 9,7 = 60,3 m
A energia potencial perdida nessa descida é:
mgΔh = 70 · 10 · 60,3 = 42.210 J
Pelo teorema trabalho-energia, a energia cinética final é igual à energia potencial perdida menos o trabalho das forças dissipativas:
E_c = 42.210 − 10.710 = 31.500 J
Daí obtemos a velocidade:
½ · 70 · v² = 31.500 → v² = 900 → v = 30 m/s
Convertendo para km/h:
v = 30 · 3,6 = 108 km/h
Resposta: D
Respondeu
levou 4 min 52 s
Como pensou
The problem is from FUVEST 2027. It involves a skateboarder named Sandro Dias performing a big drop.
Sandro starts from rest at 70 m and drops to 9.70 m above the ground. His mass is 70 kg, and friction and air drag do 10,710 J of work during the first part of the path. The speed at the lowest point will be found using the work-energy theorem.
Initial potential energy is 49,000 J and final potential energy is 6,790 J. The work done by dissipative forces is being interpreted as either 10,710 J or 10.710 J.
The work done by friction and air drag is being treated as negative, since these forces remove energy. The value is interpreted as 10,710 J.
The energy balance yields 31,500 J of kinetic energy. Solving for velocity gives 30 m/s, or 108 km/h. This matches option D.
The positive value of 10,710 J was reconsidered. If friction and drag were ignored, the speed would be about 34.7 m/s or 125 km/h, which is close to option C (126 km/h). This suggests the problem may intend for that energy loss to be excluded.
Treating 10.710 J as a decimal would still yield roughly 125 km/h, not the required 126 km/h. The exact energy balance confirms the drop is 42,210 J, so the value must be interpreted as 10,710 J.
The 126 km/h option is likely a distractor. A velocity of 35 m/s converts exactly to 126 km/h, and it could arise from ignoring the energy loss or using an incorrect mass.
The transcription likely omitted the minus sign, so 10.710 J should be read as -10,710 J. The answer 30 km/h is probably a distractor for someone who forgot to convert from m/s.
The OCR text omits the minus sign, but the context of dissipative forces confirms the energy loss must be subtracted. The 47 km/h option likely comes from a wrong conversion factor such as 30 m/s multiplied by 1.6 instead of 3.6.
The problem likely references a real skateboarding drop by Sandro Dias in Porto Alegre in October 2025. The exam version may be fictional.
The numbers were chosen to yield a perfect square. After subtracting the 10,710 J energy loss from the total drop of 42,210 J, the remaining kinetic energy is exactly 31,500 J. Dividing by half the mass gives exactly 900, so the velocity squared is exactly 900 m²/s². This confirms the answer is 30 m/s, or 108 km/h.
Adding the 10,710 J as positive work would yield a speed of about 140 km/h, which is not among the options. The dot in 10.710 might indicate a decimal point rather than a thousands separator, but that would still not match any answer.
A velocity of 35 m/s would yield exactly 126 km/h. That speed requires 42,875 J of kinetic energy. Without friction and drag, the drop should be 61.25 m to reach that energy. The actual drop is only 60.3 m, so the numbers do not align.
The comma in "9,70 m" confirms the Brazilian decimal separator. The 10,710 J value is therefore 10,710 joules, not a decimal fraction.
The 10,710 J value is about 10.7 kJ, which is a reasonable energy loss for friction and drag over this path. This confirms the work must be negative.
The problem explicitly states the work is 10.710 J, with no minus sign. This is physically inconsistent because friction and air drag are always dissipative forces. The question remains why the problem presents a positive value for energy loss.
The comma in 9,70 and the dot in 10.710 are different separators, so the value is 10,710 J. The explanation will note that friction and drag are dissipative forces, so their work is negative, and the problem likely intends W = -10,710 J.
A energia mecânica não se conserva devido às forças dissipativas. Adotando o solo como referência e interpretando o valor informado (em notação brasileira, 10.710 = 10710) como a energia retirada do sistema por atrito e arrasto, o balanço de energia entre o ponto de partida e o ponto mais baixo da rampa é
\frac12 mv^2 = mg(h_i - h_f) - 10710.
Substituindo os dados (m = 70\,\mathrm{kg}, g = 10\,\mathrm{m/s^2}, h_i = 70\,\mathrm{m}, h_f = 9{,}70\,\mathrm{m}):
mg(h_i - h_f) = 700 \times 60{,}3 = 42210\,\mathrm{J},
\frac12 \times 70 \times v^2 = 42210 - 10710 = 31500 \implies v^2 = 900 \implies v = 30\,\mathrm{m/s}.
Convertendo para \mathrm{km/h}:
30 \times 3{,}6 = 108.
Resposta: D
Respondeu
levou 6 s
Como pensou
We need answer solve physics. Need parse images. Need produce explanation in Portuguese and final line "Resposta: X".
We need solve question. Let's restate.
Problem: In October 2025, skater Sandro Dias ("Mineirinho") performed largest skate drop, descending ramp mounted lateral of Centro Administrativo Fernando Ferrari, Porto Alegre. Starting from rest on platform at 70 m above solo, skater descended ramp, passing curved transition until lowest point of ramp at height 9.70 m above solo. Path ends in flat. Consider Sandro weighs 70 kg and work of friction and air drag forces was 10,710 J during first part of path (to lowest point ramp). Considering dissipative forces, what velocity in km/h at lowest point ramp?
Given g=10 m/s². Need compute energy.
Initial height H=70 m, final height h=9.70 m, mass m=70 kg. Initial kinetic zero. Work done by non-conservative forces (friction + air drag) W_nc = -10,710 J? "trabalho das forças de atrito e arrasto do ar foi de 10.710 J durante a primeira parte" Could be magnitude? Usually work of resistive forces is negative. They might say "trabalho ... foi de 10.710 J" as energy dissipated (positive magnitude) maybe they mean magnitude. Need interpret: dissipative work decreases mechanical energy. Use conservation: ΔE = W_nc = -10,710 J. Or mgh_i + 0 = mgh_f + (1/2)mv² + W_diss? Let's see.
Initial mechanical energy: m g H = 70 * 10 * 70 = 49,000 J. Final potential: m g h = 70 * 10 * 9.70 = 6,790 J. Change in potential = -42,210 J. If no friction, kinetic = 42,210 J. With dissipative work of 10,710 J (negative), kinetic = 42,210 - 10,710 = 31,500 J. Then v = sqrt(2K/m) = sqrt(2*31500/70) = sqrt(900) = 30 m/s = 108 km/h. Alternative D. That likely answer. Let's verify if if they add 10710? If W_nc = +10,710 due to some force? No friction/drag dissipate, so magnitude 10,710 J is removed. v = 108 km/h. Options include 108.
Need craft explanation. But let's be thorough. The question asks "qual a velocidade, em km/h, que Sandro alcançou no ponto mais baixo da rampa?" Need consider dissipative effects mentioned. They say "o trabalho das forças de atrito e arrasto do ar foi de 10.710 J durante a primeira parte do percurso (até o ponto mais baixo da rampa)." It might be negative work; use energy balance.
Let's calculate in detail:
- Energia potencial inicial: Ep_i = mgh_i = 70*10*70 = 49000 J.
- Energia potencial final: Ep_f = mgh_f = 70*10*9,7 = 6790 J.
- Variação de energia potencial: ΔEp = Ep_f - Ep_i = 6790 - 49000 = -42210 J, or decrease = 42,210 J.
- Energia cinética final: Ec = decrease in potential - energy dissipated by friction/air = 42,210 - 10,710 = 31,500 J.
- Alternatively: Em_i = 49,000 J; Em_f = Ec_f + mgh_f = Ec_f + 6,790. Work nonconservative = Em_f - Em_i = -10,710. Then Ec_f + 6,790 - 49,000 = -10,710 -> Ec_f = 49,000 - 6,790 - 10,710 = 31,500 J.
- v² = 2Ec/m = 2*31500/70 = 900 -> v = 30 m/s. Convert to km/h: 30*3.6 = 108 km/h.
- Answer D.
Need ensure no issue: "primeira parte do percurso (até o ponto mais baixo da rampa)" includes vertical and transition. Yes.
Potential nuance: Work done by friction and air drag "10.710 J" maybe positive? Let's consider if they ask "considerando os efeitos das forças dissipativas mencionadas" maybe work is negative. Answer 108 indeed.
Let's produce final with explanation in Portuguese. Desired overbrevity 5. Include equations.
Need final line "Resposta: D". Maybe "Resposta: D" not "Resposta: X". The instruction says "em que X é a letra da alternativa escolhida". So final line should be Resposta: D.
Let's formulate clearly:
"Da conservação de energia: E_m,i + W_nc = E_m,f, com W_nc = -10.710 J (dissipação)." Actually sign: E_m,i + W_nc = E_m,f. If W_nc = -10.710, then E_m,f = 49.000 -10.710 = 38.290 J. Wait final mechanical energy includes final kinetic + potential. Let's check: Em_i = mgh_i = 49,000 J. W_nc = -10,710 J. Then Em_f = 49,000 - 10,710 = 38,290 J. Em_f = mgh_f + Ec_f = 70*10*9.7 + Ec = 6,790 + Ec. Thus Ec = 31,500 J. Yes correct. Wait if Em_f=38,290, then Ec=31,500. Good. Previous equation Ec_f + 6,790 - 49,000 = -10,710 => Ec_f = 31,500. Good.
Careful: If instead we use ΔK + ΔU = W_nc. ΔK = K_f - 0; ΔU = mgh_f - mgh_i = -42,210 J. So K_f - 42,210 = -10,710 -> K_f = 31,500 J. Yes.
Then v = sqrt(2K/m) = sqrt(63,000/70) = sqrt(900)=30 m/s = 108 km/h.
Let's include note: 10.710 J = 10.710? In Portuguese, 10.710 uses dot thousands, yes 10,710 J. In equations, I can use 10.710? Maybe write 10.710 J or 10710 J. Use no ambiguity: 10.710 J means 10,710 J. But in Portuguese, decimal comma, thousands dot. I'll write 10.710 J.
Let's craft final.
Pela conservação de energia, a energia potencial perdida na descida é convertida em energia cinética, mas parte dela é dissipada pelo atrito/arrasto do ar.
Dados:
- Massa: m = 70\ \text{kg}
- Altura inicial: h_i = 70\ \text{m}
- Altura final: h_f = 9,70\ \text{m}
- Energia dissipada: W_{\text{diss}} = 10.710\ \text{J}
- g = 10\ \text{m/s}^2
Energia potencial inicial:
E_{p,i}=mgh_i=70\cdot10\cdot70=49.000\ \text{J}
Energia potencial final:
E_{p,f}=mgh_f=70\cdot10\cdot9,70=6.790\ \text{J}
Variação de energia potencial:
\Delta E_p = E_{p,f}-E_{p,i}=6.790-49.000=-42.210\ \text{J}
A energia cinética final é o que “sobrou” após a dissipação:
E_c = 42.210 - 10.710 = 31.500\ \text{J}
Como E_c=\frac12 mv^2:
31.500=\frac12\cdot70\cdot v^2
v^2=\frac{2\cdot31.500}{70}=900
v=30\ \text{m/s}
Convertendo para km/h:
30\ \text{m/s}\times3,6=108\ \text{km/h}
Resposta: D
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levou 13 s
Como pensou
**Analyzing Exam Question**
I'm currently dissecting a physics problem from a FUVEST exam. My focus is on understanding the provided text and any accompanying images to extract the necessary information for a solution.
**Calculating Final Velocity**
I've processed the initial conditions and the work done by dissipative forces. Now, I'm calculating the final velocity at the ramp's lowest point using conservation of energy principles, accounting for the energy loss.
**Verifying Calculation**
I've successfully calculated the final velocity at $30\text{ m/s}$ by applying the Work-Kinetic Energy theorem and accounting for energy losses. I'm now converting this to km/h.
Para determinar a velocidade de Sandro no ponto mais baixo da rampa, podemos aplicar o princípio da conservação da energia mecânica considerando o trabalho das forças dissipativas:
-
Variação da altura (\Delta h):
\Delta h = h_{\text{inicial}} - h_{\text{final}} = 70,0\text{ m} - 9,70\text{ m} = 60,30\text{ m}
-
Trabalho do peso (variação da energia potencial gravitacional):
W_P = m \cdot g \cdot \Delta h = 70\text{ kg} \times 10\text{ m/s}^2 \times 60,30\text{ m} = 42.210\text{ J}
-
Teorema da Energia Cinética:
Como Sandro partiu do repouso (E_{c,\text{inicial}} = 0):
W_{\text{total}} = \Delta E_c
W_P + W_{\text{diss}} = E_{c,\text{final}}
42.210 - 10.710 = \frac{m \cdot v^2}{2}
31.500 = \frac{70 \cdot v^2}{2}
31.500 = 35 \cdot v^2
v^2 = \frac{31.500}{35} = 900
v = \sqrt{900} = 30\text{ m/s}
-
Conversão da velocidade para \text{km/h}:
v = 30 \times 3,6 = 108\text{ km/h}
Resposta: D
Respondeu
levou 21 s
Como pensou
**Analyzing FUVEST Physics**
I'm now focused on extracting all relevant numerical and descriptive data from the provided FUVEST 2027 physics problem, including details about the skater, Sandro Dias, and his initial velocity.
**Calculating Sandro's Velocity**
I'm working on determining Sandro's speed at the ramp's lowest point. I've identified the key parameters: initial height, final height, mass, gravity, and the work done by dissipative forces. I'm now applying the non-conservative energy relation to solve for his final velocity.
Para determinar a velocidade alcançada pelo skatista no ponto mais baixo da rampa, podemos utilizar o teorema da conservação da energia mecânica considerando a atuação de forças dissipativas (atrito e resistência do ar):
E_{m,\text{final}} = E_{m,\text{inicial}} - |W_{\text{dissipativo}}|
Adotando o ponto mais baixo da rampa (h_{\text{final}} = 9,70\text{ m}) como o nível de referência para a energia potencial gravitacional, o desnível (\Delta h) sofrido pelo skatista desde o ponto de partida (h_{\text{inicial}} = 70\text{ m}) é de:
\Delta h = 70,0\text{ m} - 9,70\text{ m} = 60,3\text{ m}
Como o skatista parte do repouso (v_{\text{inicial}} = 0), toda a sua energia mecânica inicial em relação ao ponto mais baixo é potencial gravitacional:
E_{m,\text{inicial}} = m \cdot g \cdot \Delta h
E_{m,\text{inicial}} = 70\text{ kg} \cdot 10\text{ m/s}^2 \cdot 60,3\text{ m} = 42.210\text{ J}
No ponto mais baixo da rampa, a energia mecânica resume-se à energia cinética (E_{c,\text{final}}). Subtraindo a energia dissipada pelo atrito e pelo arrasto do ar (10.710\text{ J}), temos:
E_{c,\text{final}} = 42.210\text{ J} - 10.710\text{ J} = 31.500\text{ J}
Sabendo que a energia cinética é dada por E_c = \frac{m \cdot v^2}{2}, podemos calcular a velocidade v no ponto mais baixo:
\frac{70 \cdot v^2}{2} = 31.500
35 \cdot v^2 = 31.500
v^2 = \frac{31.500}{35}
v^2 = 900 \implies v = \sqrt{900} = 30\text{ m/s}
Por fim, convertendo a velocidade de \text{m/s} para \text{km/h} (multiplicando por 3,6):
v = 30 \cdot 3,6 = 108\text{ km/h}
Resposta: D
