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Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j?
Most Upvoted Answer
Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m alon...
Understanding the Problem
To calculate the work done in pushing a block up an incline at a constant speed, we need to consider the gravitational potential energy that needs to be overcome.
Key Variables
- Mass of the block (m): 11.7 kg
- Distance along the incline (s): 4.65 m
- Height raised (h): 2.86 m
- Acceleration due to gravity (g): 9.8 m/s²
Work Done Against Gravity
When pushing the block up the incline, the work done (W) against gravity can be calculated using the formula:
W = m * g * h
This represents the gravitational potential energy gained by the block.
Calculating Work Done
Using the values:
- m = 11.7 kg
- g = 9.8 m/s²
- h = 2.86 m
The work done is:
W = 11.7 kg * 9.8 m/s² * 2.86 m = 328.23 J (approximately)
Your Approach
Your method of using the work-energy principle (WF = (Kf - Ki)) is appropriate, but in this case, since the block is moving at a constant speed, the kinetic energy does not change. Therefore, it's more straightforward to directly calculate the work done against gravity.
Conclusion
Thus, the correct work done in pushing the block up the incline at constant speed is approximately 328 J, aligning with your calculation. Remember, this work compensates for the gravitational force acting on the block as it is raised.
To calculate the work done in pushing a block up an incline at a constant speed, we need to consider the gravitational potential energy that needs to be overcome.
Key Variables
- Mass of the block (m): 11.7 kg
- Distance along the incline (s): 4.65 m
- Height raised (h): 2.86 m
- Acceleration due to gravity (g): 9.8 m/s²
Work Done Against Gravity
When pushing the block up the incline, the work done (W) against gravity can be calculated using the formula:
W = m * g * h
This represents the gravitational potential energy gained by the block.
Calculating Work Done
Using the values:
- m = 11.7 kg
- g = 9.8 m/s²
- h = 2.86 m
The work done is:
W = 11.7 kg * 9.8 m/s² * 2.86 m = 328.23 J (approximately)
Your Approach
Your method of using the work-energy principle (WF = (Kf - Ki)) is appropriate, but in this case, since the block is moving at a constant speed, the kinetic energy does not change. Therefore, it's more straightforward to directly calculate the work done against gravity.
Conclusion
Thus, the correct work done in pushing the block up the incline at constant speed is approximately 328 J, aligning with your calculation. Remember, this work compensates for the gravitational force acting on the block as it is raised.
Community Answer
Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m alon...
Block of mass m-11. 7kg
distance of-4. 65m
7.05ng
distance of-4. 65m
7.05ng
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Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j?
Question Description
Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j?.
Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j?.
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Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j?, a detailed solution for Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j? has been provided alongside types of Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j? theory, EduRev gives you an
ample number of questions to practice Block of mass m =11.7 kg is to be pushed a distance of s = 4.65 m along an incline and raised to a distance of h = 2.86 m. Assuming frictionless surface. Calculate the work done in applying a force parallel to the incline to push the block up at a constant speed. (Take, g = 9.8 ms ) = Can i solve this question like this? WF =(Kf uf )- (ki ui) =0 mgh-0 0 Wf=mgh WF =11.7x98x 2-06 = 328j? tests, examples and also practice Class 11 tests.
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