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The displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by x=t³/3 (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.?
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The displacement x of a body of mass 1 kg on smooth horizontal surface...
Calculation of Work Done for the First One Second
- Given displacement function: x = t³/3
- To find work done in the first one second, we need to calculate the change in kinetic energy of the body.
Step 1: Determining Initial and Final Positions
- Initial position (t=0): x₀ = 0
- Final position (t=1): x₁ = (1)³/3 = 1/3 m
Step 2: Calculating Work Done
- Work done = Change in kinetic energy
- Change in kinetic energy = 1/2 * m * (v₁² - v₀²)
- Since it's a smooth surface, no external forces are acting. Therefore, work done = change in kinetic energy.
Step 3: Determining Initial and Final Velocities
- Initial velocity (t=0): v₀ = dx/dt = d(t³/3)/dt = t²
- Final velocity (t=1): v₁ = dx/dt = d(t³/3)/dt = (1)² = 1
Step 4: Substituting Values to Calculate Work Done
- Work done = 1/2 * 1 * (1² - 0²) = 1/2 J
Therefore, the work done by the external agent for the first one second is 0.5 Joules. This result is obtained by calculating the change in kinetic energy of the body using the given displacement function and the concept of work-energy theorem.
- Given displacement function: x = t³/3
- To find work done in the first one second, we need to calculate the change in kinetic energy of the body.
Step 1: Determining Initial and Final Positions
- Initial position (t=0): x₀ = 0
- Final position (t=1): x₁ = (1)³/3 = 1/3 m
Step 2: Calculating Work Done
- Work done = Change in kinetic energy
- Change in kinetic energy = 1/2 * m * (v₁² - v₀²)
- Since it's a smooth surface, no external forces are acting. Therefore, work done = change in kinetic energy.
Step 3: Determining Initial and Final Velocities
- Initial velocity (t=0): v₀ = dx/dt = d(t³/3)/dt = t²
- Final velocity (t=1): v₁ = dx/dt = d(t³/3)/dt = (1)² = 1
Step 4: Substituting Values to Calculate Work Done
- Work done = 1/2 * 1 * (1² - 0²) = 1/2 J
Therefore, the work done by the external agent for the first one second is 0.5 Joules. This result is obtained by calculating the change in kinetic energy of the body using the given displacement function and the concept of work-energy theorem.
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The displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by x=t³/3 (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by x=t³/3 (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by x=t³/3 (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.?.
The displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by x=t³/3 (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by x=t³/3 (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The displacement x of a body of mass 1 kg on smooth horizontal surface as a function of time t is given by x=t³/3 (where x is in metres and t is in seconds). Find the work done by the external agent for the first one second.?.
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