Applied force in horizontal direction on block on a inclined plane.

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Homework Help Overview

The discussion revolves around a physics problem involving a 10kg block on a frictionless inclined plane at an angle of 32 degrees, subjected to a horizontal applied force of 120N. The participants are exploring how to calculate the acceleration of the block based on the forces acting on it.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the applied horizontal force and its components along the inclined plane. Questions arise regarding the method of resolving the force into components and the implications of the angle of the incline.

Discussion Status

There is an ongoing exploration of the correct approach to resolve the horizontal force into components. Some participants are questioning the initial assumptions made about the force's components, while others are reflecting on the implications of these components for calculating acceleration.

Contextual Notes

Participants are considering the effects of the angle of the incline and the nature of the forces acting on the block, including gravitational force and the applied horizontal force. The discussion highlights the need for clarity in resolving forces in a two-dimensional context.

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Homework Statement


10kg block on a inclined plane (32degrees) (frictionless) and an applied force 120N horizontally. What is acceleration?


Homework Equations



Fnet = ma

The Attempt at a Solution



mgsin32 = 51.9N (from mg)
From applied force, in the same plane, we have xcos32=120 therefore, x = 141.5
141.5-51.9=ma
a=9m/s^2 but the answer is 5.0m/s^2
 
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get_physical said:
... From applied force, in the same plane, we have xcos32=120 therefore, x = 141.5...

Explain how you got xcos32=120
 
because the horizontal force is 120N, so I need to find the hypotenuse component. the angle in that triangle is also 32, so xcos32 = 120
 
Think carefully.

The horizontal force of 120N has a component along the plane and another component perpendicular to the plane and you need to find the component along the plane.
 
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Let me think...
 
Remember that when a force is resolved into components, the force itself will be what you are calling the 'hypotenuse'.
 
Oh I get it now! Thank you!
 

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