A bus with a mass of 5 500 kg travels at a speed of 75 km/h. 1.1.1 Calculate the kinetic energy of the bus. 1.1.2 Calculate the force required when the bus brakes to come to a state of rest over 55 m.

Physics

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Answers

2 parts
1

1.19×106 J1.19 \times 10^6\ \text{J}

2

2.17×104 N2.17 \times 10^4\ \text{N}

Step-by-step solution

5 steps

Part 1· 2 steps

1.1.1 Calculate the kinetic energy of the bus.

  1. Convert speed from km/h to m/s.

    The given speed is v=75 km/hv = 75\ \text{km/h}.
    To convert to m/s, we multiply by 1000 m1 km\frac{1000\ \text{m}}{1\ \text{km}} and 1 h3600 s\frac{1\ \text{h}}{3600\ \text{s}}.

    v=75 km/h×1000 m1 km×1 h3600 sv = 75\ \text{km/h} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}} v=75×10003600 m/sv = \frac{75 \times 1000}{3600}\ \text{m/s} v=20.833... m/sv = 20.833...\ \text{m/s} v≈20.83 m/sv \approx 20.83\ \text{m/s}

  2. Calculate the kinetic energy.

    The formula for kinetic energy (KEKE) is KE=12mv2KE = \frac{1}{2}mv^2.
    Given mass m=5500 kgm = 5500\ \text{kg}.

    KE=12×5500 kg×(20.833...)2 m2/s2KE = \frac{1}{2} \times 5500\ \text{kg} \times (20.833...)^2\ \text{m}^2/\text{s}^2 KE=2750 kg×434.027... m2/s2KE = 2750\ \text{kg} \times 434.027...\ \text{m}^2/\text{s}^2 KE=1193576.38... JKE = 1193576.38...\ \text{J} KE≈1.19×106 JKE \approx 1.19 \times 10^6\ \text{J}

    The kinetic energy of the bus is 1.19 \times 10^6\ J.

Part 2

2.17×104 N2.17 \times 10^4\ \text{N}

1.1.2 Calculate the force required when the bus brakes to come to a state of rest over 55 m.

  1. Identify known values.

    Initial speed u=20.833... m/su = 20.833...\ \text{m/s} (from 1.1.1)
    Final speed v=0 m/sv = 0\ \text{m/s} (comes to rest)
    Distance s=55 ms = 55\ \text{m}
    Mass m=5500 kgm = 5500\ \text{kg}

  2. Calculate the acceleration (aa) using a kinematic equation.

    We use the equation v2=u2+2asv^2 = u^2 + 2as.

    02=(20.833...)2+2×a×550^2 = (20.833...)^2 + 2 \times a \times 55 0=434.027...+110a0 = 434.027... + 110a −110a=434.027...-110a = 434.027... a=−434.027...110a = -\frac{434.027...}{110} a=−3.9457... m/s2a = -3.9457...\ \text{m/s}^2

    The negative sign indicates deceleration.

  3. Calculate the braking force (FF).

    Using Newton's Second Law, F=maF = ma.

    F=5500 kg×(−3.9457...) m/s2F = 5500\ \text{kg} \times (-3.9457...)\ \text{m/s}^2 F=−21701.38... NF = -21701.38...\ \text{N} F≈−2.17×104 NF \approx -2.17 \times 10^4\ \text{N}

    The magnitude of the force required is 2.17 \times 10^4\ N.

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