Ella asked in Science & MathematicsPhysics · 3 months ago

A 1500kg car is coasting on a horizontal road with a speed of 18 m/s . ?

After passing over an unpaved, sandy stretch 25.0 m long, car's speed has decreased to 14 m/s . Find the magnitude of the average net force on the car in the sandy section of the road.

3 Answers

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  • 3 months ago

    To slow down the car goes through a deceleration.

    We have u = 18, v = 14 and s = 25

    Then, using v² = u² + 2as we get:

    14² = 18² + 2a(25)

    so, a = (14² - 18²)/50

    Hence, a = -2.56 ms⁻²

    Now, F = ma

    Then, F = 1500(-2.56) => -3840 Newtons

    Note, work done is force x distance

    And work done is change in kinetic energy

    so, F(25) = (1/2)(1500)(14² - 18²) 

    Hence, F = (1/2)(60)(14² - 18²)

    i.e. F = 30(14² - 18²) => -3840 N

    :)>

  • oubaas
    Lv 7
    3 months ago

    ΔKE = m/2 (14^2-18^2) = 750*-128 = -96,000 joule = F*d

    friction force F = Δ KE /d = -96/25 = -3.84 kN

  • 3 months ago

    average speed on the sandy section

    v_av = (18+14)/2 = 32/2 = 16 m/s

    time spent on sandy section

    Δt = s / v_av = 25 / 16 = 1.56 s

    average deceleration

    a = Δv/Δt

    Δv = speed after - speed before

    a = (14-18) / 1.56 = -4 / 1.56 = -2.56 m/s²

    the magnitude of the average net force

    F = m a

    F = 1500 * (-2.56) = -3840N

    "-" means it's a braking force 

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