Recent content by thatstheguy9

I see where I went wrong. Thanks for your help @haruspex and @Lnewqban

How can they have the same angular velocity when they have different radi? Which brings me to my next point of confusion: Why they have used the radius of pulley D to calculate the angular velocity of pulley B.

I got 75 rad/s using: ## V_D = V_B ## ## \omega_D r_D = \omega_B r_B ## ## 300*0.025 = \omega_B * 0.100 ## ## \omega_B = 75 rad/s##

Pulley B and D

Apologies, I've corrected the original post.

So far I have: The velocity of the belt will be the same for pully A and D, so we can calculate the angular velocity of pulley D: ## V_A = V_B ## ## \omega_A r_A = \omega_D r_D ## ## ((20*3)+40)(0.075) = \omega_D (0.025) ## ## \omega_D = 300 Rad/s ## My next step was to determine the angular...

I got 3.9 m/s

Thanks for the response @PeroK. I have already solved parts 1, 2 and 3 and used conservation of energy in two phases like you suggested for part 3. What's thrown me off the part 4. Are you saying I should apply ## F = ma ## in two phases for this question? Using energy to find the velocity as a...

Sorry @haruspex I don't recognize that. Is the second derivative of displacement with respect to time? I haven't done any form of harmonic motion so the only concepts I have to solve it are particle dynamics and conservation of energy

My interpretation of the question was that we are looking at the motion as the block is slowed by the spring. That makes sense, however I'm having struggling with deriving the velocity. So far my acceleration = -25x - 2.452 To find the velocity I am using: a dx = v dv, is this the correct...

I see. I'm unsure what to do then lol

My apologies, yes that is correct. I see. Thank you

My approach so far is to use F = ma. The forces acting on the block in the horizonital direction are friction and the force of the spring. Choosing the direction towards the spring as the positive axis. Therefore: F = ma -Fr - kx = ma Solving for a = (-Fr - kx)/m If I plug in values I end up...