Block pushed up frictionless incline

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SUMMARY

The discussion focuses on calculating the acceleration of a block pushed up a frictionless incline at a 30-degree angle with an applied force of 50 N and a mass of 3 kg. The correct acceleration is determined to be 11.77 m/s², derived from applying Newton's second law to the forces acting on the block. Key steps include resolving forces into components parallel and perpendicular to the incline and utilizing a free body diagram for clarity. The solution emphasizes the importance of considering gravitational force components in the calculations.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of vector components in physics
  • Ability to draw and interpret free body diagrams
  • Familiarity with trigonometric functions, particularly sine and cosine
NEXT STEPS
  • Study the application of Newton's second law in various contexts
  • Learn how to effectively draw and analyze free body diagrams
  • Explore the effects of different angles on acceleration in inclined planes
  • Investigate the role of normal force in inclined plane problems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding dynamics on inclined planes, particularly in scenarios involving frictionless surfaces and force analysis.

mooneh
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a block is pushed up a frictionless 30 incline by an applied force
the force F=50 N and the mass = 3 kg
whats the magnitude of the resulting acceleration of the block ?


sol:
-Fcos30=ma
mg+Fsin30=ma

right answer for a=9.4
 
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These problems will be easier to solve if:
(1) You draw yourself a free body diagram
(2) You use components parallel (and perpendicular) to the plane

In this case, apply Newton's 2nd law to force components parallel to the plane (since that's the only direction that the object can move).

(Extra hint: If you insist on using vertical and horizontal components, you will need to add the normal force into the mix.)
 
Last edited:
Right answer is a=11.77m/s^{}2
Because F-mgsin(30)=ma so 50-3*9.8*0.5=3a and a=11.77m/s^{}2
 

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