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Zorluk: OrtaMechanics and Motion

A maintenance cart with a mass of 250 kg250\text{ kg} travels along a straight horizontal track. If the cart uniformly slows down from an initial velocity of 15 m/s15\text{ m/s} due east to 5 m/s5\text{ m/s} due east over a time interval of 5 seconds5\text{ seconds}, what is the magnitude and direction of the net force acting on the cart?

  1. A
    500 N500\text{ N} directed due east
  2. 500 N500\text{ N} directed due westCevap
  3. C
    4900 N4{}900\text{ N} directed due west
  4. D
    0.5 N0.5\text{ N} directed due west

Cevap

The magnitude of the net force is 500 N500\text{ N}, directed due west.
Acceleration is the change in velocity divided by time: a=(5 m/s−15 m/s)/5 s=−2 m/s2a = (5\text{ m/s} - 15\text{ m/s}) / 5\text{ s} = -2\text{ m/s}^2. The negative sign indicates an acceleration directed due west (opposite to the eastward motion). Applying Newton's Second Law (F=maF = ma), the net force is 250 kg×2 m/s2=500 N250\text{ kg} \times 2\text{ m/s}^2 = 500\text{ N} directed due west.

Adım Adım Çözüm

1
Calculate the acceleration using the kinematics formula a=vf−viΔta = \frac{v_f - v_i}{\Delta t}.
a=5 m/s−15 m/s5 s=−10 m/s5 s=−2 m/s2a = \frac{5\text{ m/s} - 15\text{ m/s}}{5\text{ s}} = \frac{-10\text{ m/s}}{5\text{ s}} = -2\text{ m/s}^2 (or 2 m/s22\text{ m/s}^2 directed west).
Acceleration is the rate of change of velocity over time.
2
Calculate the magnitude and direction of the net force using Newton's Second Law, Fnet=m⋅aF_{\text{net}} = m \cdot a.
Fnet=250 kg×2 m/s2=500 NF_{\text{net}} = 250\text{ kg} \times 2\text{ m/s}^2 = 500\text{ N} directed west.
Net force is equal to mass multiplied by acceleration and always points in the same direction as the acceleration vector.

Anahtar Kavram

Newton's Second Law of Motion (Fnet=maF_{\text{net}} = ma) and Kinematics of Linear Deceleration

Alternatif Yöntem

Use the impulse-momentum theorem (Fnet⋅Δt=m⋅ΔvF_{\text{net}} \cdot \Delta t = m \cdot \Delta v). Here, Δp=250 kg×(5 m/s−15 m/s)=−2500 kg⋅m/s\Delta p = 250\text{ kg} \times (5\text{ m/s} - 15\text{ m/s}) = -2{}500\text{ kg}\cdot\text{m/s}. Dividing momentum change by Δt=5 s\Delta t = 5\text{ s} yields Fnet=−500 NF_{\text{net}} = -500\text{ N}, meaning 500 N500\text{ N} directed west.
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Mechanics and Motion

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