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What are the direct values of forests?

Mathematics

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4 parts
a

5−3 m/s25 - \sqrt{3}\ \mathrm{m/s^2}

b

5,11 m/s5,11\ \mathrm{m/s}

c

−83 J-8\sqrt{3}\ \mathrm{J}

d

40−83 J40 - 8\sqrt{3}\ \mathrm{J}

Step-by-step solution

11 steps

Part a· 4 steps

A aceleração do bloco

  1. A aceleração é dada pela fórmula

    a=g(sin⁡θ−μcos⁡θ)a = g(\sin\theta - \mu\cos\theta)

    onde g=10 m/s2g = 10\ \mathrm{m/s^2}, θ=30∘\theta = 30^\circ, μ=0,2\mu = 0,2.

  2. Calcular sin⁡θ\sin\theta e cos⁡θ\cos\theta

    sin⁡30∘=12, cos⁡30∘=32\sin 30^\circ = \frac{1}{2},\ \cos 30^\circ = \frac{\sqrt{3}}{2}

  3. Substituir

    sin⁡θ−μcos⁡θ=12−0,2×32=12−0,13\sin\theta - \mu\cos\theta = \frac{1}{2} - 0,2 \times \frac{\sqrt{3}}{2} = \frac{1}{2} - 0,1\sqrt{3}

  4. Simplificar

    a=10(12−0,13)=5−3 m/s2≈3,27 m/s2a = 10\left(\frac{1}{2} - 0,1\sqrt{3}\right) = 5 - \sqrt{3}\ \mathrm{m/s^2} \approx 3,27\ \mathrm{m/s^2} 5−3 m/s2\boxed{5 - \sqrt{3}\ \mathrm{m/s^2}}

Part b

5,11 m/s5,11\ \mathrm{m/s}

A velocidade do bloco em B

  1. Usando v2=u2+2asv^2 = u^2 + 2as com u=0u = 0, s=L=4 ms = L = 4\ \mathrm{m}

    v2=2aLv^2 = 2aL v=2aLv = \sqrt{2aL}

  2. Substituir a=5−3 m/s2a = 5 - \sqrt{3}\ \mathrm{m/s^2}, L=4 mL = 4\ \mathrm{m}

    2aL=2(5−3)×4=8(5−3)=40−832aL = 2(5 - \sqrt{3}) \times 4 = 8(5 - \sqrt{3}) = 40 - 8\sqrt{3}

  3. Simplificar

    v=40−83 m/s≈5,11 m/sv = \sqrt{40 - 8\sqrt{3}}\ \mathrm{m/s} \approx 5,11\ \mathrm{m/s} 5,11 m/s\boxed{5,11\ \mathrm{m/s}}

Part c

−83 J-8\sqrt{3}\ \mathrm{J}

O trabalho realizado pelo atrito

  1. Força de atrito f=μmgcos⁡θf = \mu mg \cos\theta

    f=0,2×2×10×32=4×32=23 Nf = 0,2 \times 2 \times 10 \times \frac{\sqrt{3}}{2} = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}\ \mathrm{N}

  2. Trabalho Wa=−fLW_\mathrm{a} = -fL (sentido oposto ao deslocamento)

    Wa=−23×4=−83 J≈−13,86 JW_\mathrm{a} = -2\sqrt{3} \times 4 = -8\sqrt{3}\ \mathrm{J} \approx -13,86\ \mathrm{J} −83 J\boxed{-8\sqrt{3}\ \mathrm{J}}

Part d

40−83 J40 - 8\sqrt{3}\ \mathrm{J}

A energia cinética do bloco em B

  1. Ec=12mv2E_\mathrm{c} = \frac{1}{2}mv^2

  2. Como m=2 kgm = 2\ \mathrm{kg}, Ec=v2=40−83 J≈26,14 JE_\mathrm{c} = v^2 = 40 - 8\sqrt{3}\ \mathrm{J} \approx 26,14\ \mathrm{J}

    Alternativa (energia mecânica): Ec=mgh+WaE_\mathrm{c} = mgh + W_\mathrm{a}, h=Lsin⁡θ=4×0,5=2 mh = L\sin\theta = 4 \times 0,5 = 2\ \mathrm{m}

    mgh=2×10×2=40 Jmgh = 2 \times 10 \times 2 = 40\ \mathrm{J} Ec=40+(−83)=40−83 JE_\mathrm{c} = 40 + (-8\sqrt{3}) = 40 - 8\sqrt{3}\ \mathrm{J} 40−83 J\boxed{40 - 8\sqrt{3}\ \mathrm{J}}
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