A 10 kg object is moving with a speed of 3 m/s. If it comes to a stop, how much work is done by the friction force?

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Q: A 10 kg object is moving with a speed of 3 m/s. If it comes to a stop, how much work is done by the friction force?

Solution: Work done = Change in Kinetic Energy = 0 - (0.5 × 10 kg × (3 m/s)²) = -45 J.

Steps: 10

Step 1: Identify the mass of the object. The mass is 10 kg.
Step 2: Identify the initial speed of the object. The speed is 3 m/s.
Step 3: Calculate the initial kinetic energy using the formula: Kinetic Energy = 0.5 × mass × (speed)².
Step 4: Plug in the values: Kinetic Energy = 0.5 × 10 kg × (3 m/s)².
Step 5: Calculate (3 m/s)², which is 9 m²/s².
Step 6: Multiply: 0.5 × 10 kg × 9 m²/s² = 45 J. This is the initial kinetic energy.
Step 7: When the object comes to a stop, its final kinetic energy is 0 J.
Step 8: Calculate the change in kinetic energy: Change in Kinetic Energy = Final Kinetic Energy - Initial Kinetic Energy.
Step 9: Substitute the values: Change in Kinetic Energy = 0 J - 45 J.
Step 10: The change in kinetic energy is -45 J, which means the work done by the friction force is -45 J.