A box of mass 8.0 kg slides at a speed of 10 ms⁻¹ across a smooth level floor before it encounters a rough patch of length 3.0 m. The frictional force on the box due to this part of the floor is 70 N. What is the speed of the box when it leaves this rough surface? What length of the rough surface would bring the box completely to rest?

Introduction

This problem deals with the effect of friction on a moving object. When a box moves across a rough surface, the frictional force does negative work, reducing the speed of the box. We’ll apply the work-energy theorem to solve the two parts of this problem.

Given Data

  • Mass of box, m = 8.0 kg
  • Initial speed, u = 10 m/s
  • Frictional force, f = 70 N
  • Length of rough surface, s = 3.0 m
  • We are asked to find:
    • (a) Final speed of the box after covering 3.0 m
    • (b) Distance required to bring the box to rest

Step 1: Work-Energy Theorem

Work done by friction = Change in kinetic energy

Work done = -f × s (negative because friction opposes motion)

Initial K.E. = (1/2)mu² = 0.5 × 8 × (10)² = 400 J

Work done = -70 × 3 = -210 J

Change in K.E. = Final K.E. – Initial K.E. = Work done

Final K.E. = 400 – 210 = 190 J

Final speed, v is given by:

½mv² = 190 → v² = (2 × 190)/8 = 47.5 → v = √47.5 ≈ 6.89 m/s

Step 2: Distance to Bring the Box to Rest

We now want the box to stop completely (final velocity = 0).

Initial K.E. = ½ × 8 × 10² = 400 J

To bring the box to rest, all this energy must be removed by friction:

Work done by friction = -f × s = -70 × s

So, -70 × s = -400 → s = 400 / 70 ≈ 5.71 m

Final Answers

  • Speed of box after 3 m: 6.89 m/s
  • Distance required to stop box: 5.71 m

Conclusion

This problem demonstrates how friction reduces the motion of an object. By using energy concepts, we determined that the speed of the box reduces to approximately 6.89 m/s after 3 m, and it would take around 5.71 m of rough surface to bring it to a complete stop.

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