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That’s only F net in the “y” direction which in this problem is perpendicular to the surface of the ramp. All this means is that the box doesn’t fly off of the ramp or sink down into the ramp. So, the y components of the force due to gravity, normal force, and vertical pulling force will cancel out because there is no acceleration in the y direction. There is still a force on the box in the “x” direction.

Glad I could help!

If you are talking about what grade as in high school levels, this is usually either a senior level high school class (12th grade) or a freshman level college class.

@ZeroWaitTutoring OK thanks 😊

Because the differential equation is nonlinear, I don’t see another method to solve. Splitting the dx and dy is a legitimate method to solving separable differential equations.

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Work is equal to the dot product of Force and displacement. The definition of the dot product states that work is equal to the force times the displacement, times the angle between the force and displacement. In this problem, delta X is up the ramp and the force of friction is down the ramp. The angle between the displacement and frictional force is 180 degrees.

For this problem it is most convenient to set the x-axis as the sloped part of the ramp. This makes the y-axis perpendicular to the x-axis (perpendicular to the ramp’s surface). For the forces involved in this problem, trigonometry is used to split the forces into their components in these new x and y axis.

@@ZeroWaitTutoring Alright, thanks!

Yes, still the same process. Since all of the charges are reversed, the direction would still be down and to the left.

The distance values given from the sun/moon to the earth are measured from the center of the sun/moon to the center of the earth. The distance from the center of the earth to point b is re. So the distance from the sun/moon to point b of the earth is d+re. Also yes, diameter is 2r.

@@ZeroWaitTutoring ahhhh! Thank you

Make sure you are following the order of operations. Do (sin30-.26cos30) you will get ~.275 (make sure your calculator is in degree mode). Then multiply what’s outside the parentheses, mg, which is (9.8 * 3.2) = 31.36. Multiply the two numbers in the numerator ( .275 * 31.36) = 8.624. Now in the denominator do (mass of the pulley/2) = (1.8/2) = .9, then add it to m. You’ll get (.9+3.2) = 4.1. So, your numerator is 8.624 and your denominator is 4.1, (8.624/4.1) = 2.103 .

You would need to modify the conservation of momentum equation to account for V2 being negative as well as the change of momentum being negative. it would be: -(change of momentum) = -mV2 - mV1. with numbers: -1.4 = -.14V2 - .14(5.2). Then, -1.4 + .14(5.2) = -.14V2. Then -.14V2 = -.672. Divide -.14 to both sides and get V2 = 4.8 m/s. Since you specified that V2 was negative (by saying that V1 is +), you would add a negative in the front or say that it's to the left.

.55 rev/s is a measure of frequency. My equation contained period which has units of s/rev. So, I needed to convert rev/s to s/rev in order to use it in my equation.