Work and energy with normal force

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Homework Help Overview

The discussion revolves around a physics problem involving work and energy, specifically focusing on the work done by the normal force on a block sliding down an incline. The scenario includes parameters such as mass, incline angle, coefficient of friction, and gravitational acceleration.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand how to calculate the work done by the normal force and questions how to determine the angle associated with it. Some participants clarify the orientation of the normal force in relation to the incline and discuss the implications of its perpendicular nature on work done.

Discussion Status

Participants are exploring the relationship between the normal force and the work done, with some suggesting that the work may be zero due to the normal force acting perpendicular to the motion. There is an ongoing examination of the role of friction in this context, indicating a productive dialogue about the underlying concepts.

Contextual Notes

There is a mention of potential confusion regarding the angle of the incline and its relation to the normal force, as well as the need to consider the effects of friction in the analysis of work done.

Maiia
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Homework Statement


Starting from rest, a 5.85 kg block slides 2.07m down a rough 26.4degree incline. The coefficient of kinetic friction between the block and the incline is 0.427. The acceleration of gravity is 9.8m/s^2. Find the work done by the normal force. Answer in units of Joules.

Here's my FBD.
physics.jpg


I know the formula for work is Fcos(theta)d. But I don't know how to find the angle for the normal force...
If someone could help out, I would really appreciate it.
 
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The normal force acts at right angles to the slope.
You 26.4 is in the wrong place - a slope angle of 26.4 means the angle between the slope and horizontal.
 
yes, but if you draw some parallel lines, the 26.4 degrees of the slope angle is also equivalent to that angle up there. oh but i see what you said about the normal being perpendicular...wouldnt that mean work is zero then?
 
Maiia said:
yes, but if you draw some parallel lines, the 26.4 degrees of the slope angle is also equivalent to that angle up there. oh but i see what you said about the normal being perpendicular...wouldnt that mean work is zero then?

Yes you are right that forces in the direction of motion are the ones that do work, but in the case of the Normal force there is a resulting force along the direction of motion as determined by the Normal force * the Coefficient of Friction.
 

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