Gravity Equation: Prob 3. Solving Force in Inclined Plane

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The discussion revolves around calculating the force required to push a 20 kg block up a 37-degree inclined plane with a constant speed of 2.5 m/s, considering a kinetic friction coefficient of 0.25. The force F is derived using the equation F = 0.25*200cos37 + 200sin37, where 200sin37 represents the gravitational force component acting parallel to the incline. Participants clarify that the gravitational force is indeed calculated as Wg = -20*200sin37, highlighting the importance of understanding the components of gravitational force in relation to the incline. The conversation emphasizes distinguishing between gravitational potential energy and the gravitational force acting parallel to the ramp. Overall, the focus is on accurately breaking down the forces involved in the inclined plane scenario.
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Homework Statement



Prob. 3. A block of mass M = 20.0 kg is pushed up a rough inclined plane 37o to the horizontal by a constant force F parallel to the incline as shown. The co-efficient of kinetic friction between the block and the incline is µ = 0.25, and the block is moving up with a constant speed of 2.5 m/s between A and B.

Homework Equations





The Attempt at a Solution



What is the magnitude of F ?
F = 0.25*200cos37 + 200sin37 = 160N
Can you explain the 200sin37 part. Is that gravity can you break it apart
also
gravitational force ? Wg = - 20*200sin37 = - 2400J :
i thought it would be 20*10*20sin37
is gravity force g*m*d where d is the hypotenuse or is the vertical side of the right triangle?

thanks
 
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You're thinking about gravitational potential energy. 200sin37 is the component of gravity parallel to the ramp.
 
what about the gravitation force?
- 20*200sin37

is that parallel to the ramp as well?
 

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