Question 16 of 19advanced🔧 ApplyNumerical5 marks

An object of mass 2 kg moving with a constant velocity of 10 m s⁻¹ encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?

Correct Answer

10 m

Exercise: Revise, Reflect, Refine | Q: 14 | (Page 21)
For More Understanding

Explanation

The problem requires calculating the net retarding force, then the deceleration using Newton's second law, and finally the distance using the kinematic equation v2v^{2} = u2u^{2} + 2as. The textbook context provides examples of calculating net force and acceleration (Example 6.5) and using kinematic equations (Example 6.6).

Solution Steps

  1. Step 1: Calculate the net opposing force: F = 7 N + 3 N = 10 N

  2. Step 2: Calculate the acceleration using F = ma: a = F/m = 10 N / 2 kg = 5 m/s². Since it opposes motion, a = -5 m/s²

  3. Step 3: Use the kinematic equation v² = u² + 2as to find distance: 0 = (10)² + 2(-5)s

  4. Step 4: Solve for s: 0 = 100 - 10s => 10s = 100 => s = 10 m

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