Examkin

Question

Difficulty: MediumMechanics and Motion

An electric delivery van with a total mass of 2,000 kg2,000\text{ kg} travels along a straight road at a constant velocity of 15 m/s15\text{ m/s}. The driver applies the brakes, producing a constant opposing friction force of 6,000 N6,000\text{ N} until the vehicle comes to a complete stop. How much time does it take for the van to stop?

  1. A
    20 seconds20\text{ seconds}
  2. 5 seconds5\text{ seconds}Answer
  3. C
    50 seconds50\text{ seconds}
  4. D
    45 seconds45\text{ seconds}

Answer

The total time required for the delivery van to come to a complete stop is 5 seconds5\text{ seconds}.
According to Newton's second law (F=maF = ma), the opposing braking force of 6,000 N6,000\text{ N} acting on a 2,000 kg2,000\text{ kg} vehicle produces an acceleration of a=6,000 N2,000 kg=3 m/s2a = \frac{6,000\text{ N}}{2,000\text{ kg}} = 3\text{ m/s}^2. To bring the vehicle from 15 m/s15\text{ m/s} to rest at a rate of 3 m/s23\text{ m/s}^2, the time needed is t=Δva=15 m/s3 m/s2=5 secondst = \frac{\Delta v}{a} = \frac{15\text{ m/s}}{3\text{ m/s}^2} = 5\text{ seconds}.

Step-by-Step Solution

1
Calculate the magnitude of deceleration using Newton's second law of motion (F=maF = ma).
a=Fm=6,000 N2,000 kg=3 m/s2a = \frac{F}{m} = \frac{6,000\text{ N}}{2,000\text{ kg}} = 3\text{ m/s}^2
The net opposing force applied to the known mass determines the rate at which the vehicle slows down.
2
Determine the time required to reduce the velocity from 15 m/s15\text{ m/s} to 0 m/s0\text{ m/s} using the kinematic definition of acceleration (t=Δvat = \frac{\Delta v}{a}).
t=15 m/s3 m/s2=5 secondst = \frac{15\text{ m/s}}{3\text{ m/s}^2} = 5\text{ seconds}
Uniform deceleration reduces speed at a constant rate of 3 m/s3\text{ m/s} each second until rest is reached.

Key Concept

Newton's Second Law of Motion and Linear Deceleration
Estimated Time:1m 0s
Rate this question

Topics

Mechanics and Motion

More questions from this topic