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Zorluk: ZorMechanics and Motion

A transport vehicle with a mass of 2000 kg2{}000\text{ kg} travels along a straight, level road at a constant velocity of 20 m/s20\text{ m/s}. The driver applies the brakes, causing the vehicle to decelerate uniformly to a complete stop over a distance of 40 m40\text{ m}. What is the magnitude of the net horizontal braking force acting on the vehicle?

  1. A
    1000 N1{}000\text{ N}
  2. 10000 N10{}000\text{ N}Cevap
  3. C
    19600 N19{}600\text{ N}
  4. D
    100000 N100{}000\text{ N}

Cevap

The magnitude of the net horizontal braking force is 10000 N10{}000\text{ N}.
The magnitude of the acceleration is determined by vf2=vi2+2adv_f^2 = v_i^2 + 2ad, yielding a=5 m/s2a = 5\text{ m/s}^2. According to Newton's Second Law (F=maF = ma), multiplying the mass of 2000 kg2{}000\text{ kg} by 5 m/s25\text{ m/s}^2 results in a net braking force of 10000 N10{}000\text{ N}.

Adım Adım Çözüm

1
Calculate the magnitude of uniform acceleration using the kinematic equation relating velocity, acceleration, and displacement: vf2=vi2+2aΔxv_f^2 = v_i^2 + 2a\Delta x.
02=(20 m/s)2+2a(40 m)  ⟹  0=400+80a  ⟹  a=−5 m/s20^2 = (20\text{ m/s})^2 + 2a(40\text{ m}) \implies 0 = 400 + 80a \implies a = -5\text{ m/s}^2. The magnitude of acceleration is 5 m/s25\text{ m/s}^2.
The acceleration must be found from the known initial speed (20 m/s20\text{ m/s}), final speed (0 m/s0\text{ m/s}), and stopping distance (40 m40\text{ m}).
2
Apply Newton's Second Law of Motion (Fnet=maF_{\text{net}} = ma) to find the magnitude of the net horizontal braking force.
Fnet=(2000 kg)(5 m/s2)=10000 NF_{\text{net}} = (2{}000\text{ kg})(5\text{ m/s}^2) = 10{}000\text{ N}.
The net force required to decelerate the mass is the product of mass and acceleration.

Anahtar Kavram

Kinematic equations for uniform deceleration and Newton's Second Law of Motion (F=maF = ma)

Alternatif Yöntem

Find the average velocity vavg=vi+vf2=20+02=10 m/sv_{\text{avg}} = \frac{v_i + v_f}{2} = \frac{20 + 0}{2} = 10\text{ m/s}. The stopping time is t=dvavg=4010=4 st = \frac{d}{v_{\text{avg}}} = \frac{40}{10} = 4\text{ s}. Acceleration is a=Δvt=204=5 m/s2a = \frac{\Delta v}{t} = \frac{20}{4} = 5\text{ m/s}^2, giving F=ma=2000×5=10000 NF = ma = 2{}000 \times 5 = 10{}000\text{ N}.
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Mechanics and Motion

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