Help finding the magnitude of the applied force

AI Thread Summary
To find the magnitude of the applied force on a 50 kg block sliding on a frictionless incline at a 20° angle with an acceleration of 4 m/s², the equation F=ma yields a force of 200 N. The confusion arises from the calculation involving the cosine of the angle, where 200 N is multiplied by cos(20°) to determine the component of the force acting parallel to the incline. The resulting force in the x-direction is approximately 187.9 N. The applied force must also counteract the component of gravitational force acting down the incline, which is where the additional 167 N comes into play, leading to a total applied force of 367 N. Understanding these components clarifies the block's motion and the forces at work on the incline.
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Homework Statement


A block of mass 50 kg is free to slide on a frictionless inclined plane that makes an angle of 20° with respect to horizontal. A force F is applied as shown in the drawing pulling parallel to the incline. As a result, the block accelerates up the plane at 4 m/s2.

known:
m= 50 kg
a= 4 m/s^2
Angle = 20 degrees


Homework Equations


F=ma
Fcos(theta)


The Attempt at a Solution


F= 50kg* 4m/s2 = 200 N
200cos(20)=187.9= F x-direction

* The explanation shows me F- 167 N = 200N
F=367N
But I don't know which equation this comes from or where the 167 comes from.
 
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3. The Attempt at a Solution
F= 50kg* 4m/s2 = 200 N
200cos(20)=187.9= F x-direction
......
200cos(20)=187.9= F x-direction
What force is this?
 
This was my attempt at finding the applied force..
 
Can you tell me why you multiply 200 with Cos(20)?

If you just put an object on frictionless inclined plane, in which direction the object moves?. What makes it move in that direction?
 
Last edited:
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