The correct answer for the question that is being presented above is this one: "(d) the magnitude of the normal force on the box from the ramp." A box is on a ramp that is at angle θ to the horizontal. As θ is increased from zero, and before the box slips, do the following increase, decrease, or remain the same and <span> the magnitude of the normal force on the box from the ramp.</span>
Answer:
We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

Only the x-component of the applied horizontal force does work on the trunk.
But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.



The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

The energy dissipated by the frictional force can be found as follows:

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.
In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.




Explanation:
As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.
Answer:
Explanation:
The magnitude of the acceleration makes an angle of 30° with the tangential velocity.
Resolving the acceleration to tangential and radial acceleration
at = aCos30 = √3a/2
ar = aSin30 = ½a
a = 2•ar
Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:
at = Rα
Then, α = at/R
since at = √3a/2
Then, α = √3 at/2R, equation 1
The radial acceleration is given as
ar = ω²R
Note that, at² + ar² = a²
at = √(a²-ar²)
Back to equation 1
α = √3 at/2R
α = √3√(a²-ar²)/2R
α = √3√(a²-(w²R)²)/2R
α = √3(a²-w⁴R²) / 2R
Also, a = 2•ar = 2w²R
Then,
α = √3((2w²R)²-w⁴R²) / 2R
α = √3(4w⁴R²-w⁴R²) / 2R
α = √3(3w⁴R²) / 2R
α = √9w⁴R² / 2R
α = 3w²R / 2R
α = 3w²/2