Recent content by Lone Wolf

Figure: a) The mechanical energy of the sphere is conserved because the weight is the only force which does work. My problem with this question is mostly because the original picture (which I tried to recreate here) is kind of ambiguous, as in I don't know if H already accounts for the radius...

a) ΔK = 1064 = 1/2 * I * (ω²(3) - ω²(2)) I = 2*1064/(ω²(3) - ω²(2)) If I assume α = n*t²: α = dω/dt --> ω = ∫ n* t² dt = n*t³/3 + ω0 = n*t³/3 ω(5) = 100 = n*5³/3 --> n = 2.4 α = 2.4 t² and ω = 2.4 * t³/3 ω(3) = 21.6 rad/s and ω(2) = 6.4 rad/s. Replacing the values: I = 2*1064/(21.6²-6.4²)...

But according to Newton's third law, wouldn't the force the man exerts on the ladder be equal to the reaction force exerted by the ladder on him?

I solved this question correctly, however I have a question regarding how I should work with the weight of the firefighter climbing the ladder. When drawing the force diagram for this problem, I should only include forces acting on the ladder, right? Which means I would represent the normal...

Let v be the speed of the block and x elongation of the spring beyond the equilibrium point. Initially, v = 0 and x = 0. At the maximum elongation, the block also has v = 0, it has moved a distance equal to x (parallel to the plane) and the variation of height is equal to -x⋅sin(53°). W(FNC) =...

Thank you for taking your time. I appreciate it!

a) Let m be the vehicle's mass, M the truck's mass, vt the truck's speed, vc the car's speed, vf the final speed, θ the angle both vehicles make with the horizontal axis (west-east direction) after the collision. Conservation of linear momentum: In the x direction: M vt = (m + M) vf cos(θ) In...

Yeah that was it. Looks like I got distracted while I was solving the problem. Thanks for spotting my mistake!

The object is: My attempt at a solution: I divided the object into 3 different rectangles and found the coordinates for the center of mass of each one, considering the origin at point "O". Then I found the mass of each rectangle, assuming the object has an area density of σ. m1 = 15σ; m2= 6σ...

a) Solution given: F = - x î - y j b) The equilibrium position happens when F = 0. x = 0 and y = 0 is the point of equilibrium. Solution given: (0, 0) c) Since the particle has a circular trajectory the trajectory equation becomes x^2 + y^2 = R^2. The maximum potential energy the...