A 10 kg object is moving with a speed of 3 m/s. If it comes to a stop, how much

₹0.0

Question: 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?

Options:

Correct Answer: 90 J

Solution:

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

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?

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?

Correct Answer: -45 J

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.

Work-Energy Principle – The work done by forces on an object is equal to the change in its kinetic energy.
Kinetic Energy Calculation – Kinetic energy is calculated using the formula KE = 0.5 × mass × velocity².
Friction Force – Friction force does negative work when it opposes the motion of an object.