In a physics experiment, the speed of an object is measured as 20.0 m/s with an uncertainty of ±0.4 m/s. What is the maximum possible error in the kinetic energy calculated from this speed?
Practice Questions
1 question
Q1
In a physics experiment, the speed of an object is measured as 20.0 m/s with an uncertainty of ±0.4 m/s. What is the maximum possible error in the kinetic energy calculated from this speed?
8 J
4 J
2 J
1 J
Kinetic energy = 0.5 * m * v²; maximum error = m * v * uncertainty in v = m * 20.0 * 0.4 = 8 J (assuming m = 1 kg).
Questions & Step-by-step Solutions
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Q
Q: In a physics experiment, the speed of an object is measured as 20.0 m/s with an uncertainty of ±0.4 m/s. What is the maximum possible error in the kinetic energy calculated from this speed?
Solution: Kinetic energy = 0.5 * m * v²; maximum error = m * v * uncertainty in v = m * 20.0 * 0.4 = 8 J (assuming m = 1 kg).
Steps: 8
Step 1: Understand the formula for kinetic energy, which is KE = 0.5 * m * v², where KE is kinetic energy, m is mass, and v is speed.
Step 2: Identify the given values: speed (v) = 20.0 m/s and uncertainty in speed = ±0.4 m/s.
Step 3: Recognize that the uncertainty in speed affects the kinetic energy calculation.
Step 4: Calculate the maximum possible error in kinetic energy using the formula: maximum error = m * v * uncertainty in v.
Step 5: Assume a mass (m) of 1 kg for simplicity in calculation.
Step 6: Substitute the values into the formula: maximum error = 1 kg * 20.0 m/s * 0.4 m/s.
Step 7: Perform the multiplication: 1 * 20.0 * 0.4 = 8 J.
Step 8: Conclude that the maximum possible error in the kinetic energy is 8 J.
