Recent content by Chemistrystrng

Oh, yes. It is coming down the slope, so it's final - initial, so 0-4.7, so it's -4.7, which would take care of the - acceleration. Giving the answer to be a positive 5.7 m/s. Thank you so much. That helped a lot.

Oh, okay so the velocity is negative, and so is the acceleration since it's speeding up. But when using the V2 equation, it comes out to be V2 = sqrt (-33.276) since initial velocity is zero (I think). Do I make the acceleration positive since it's asking for speed and not velocity? I'm slow =(

I got the answer as 5.7 m/s and it is correct. Using the 4.7 incline height. But, is the acceleration positive or negative as it goes down? Because isn't the initial velocity 0? And if so, would the signs of those two forces mentioned above be different?

Okay, so as it's going down, would the forces in the x direction be -mgsintheta + fk? and then I substitute the fk for umgcostheta and solve for acceleration, correct? same as part a except we have the 2.5 m height this time, right?

And for part b, we just use the V2 equation since we have the incline height? Or am I completely wrong again?

The distance along the incline...so you would do 4.7sin32 to get the vertical height..giving 2.5 m. Thank you so much! I thought the 4.7 was the vertical height. I have trouble with physics problems...I understand the concepts, but I can't do the actual problems without outside help. It's...

Oh, so you don't need to use 8sin32? It's just 8 m/s? If so, then the height is 4.7, if the negative sign is not squared. But that is also incorrect... By making it positive, I meant the V2 - Vi2

Wait, no i did not get that. I got -1.3 m...and I know that's not right. But I used V2 = Vi2 + 2ay...

I got 7.71 m..but I don't know if this is correct or not. I assumed final velocity was zero and I used 8sin32 to calculate the initial velocity in the direction of the object. But what I don't understand is that it's 0-8sin32^2, making it positive, and since acceleration is negative, the answer...

Hi :) It's the vertical height. I forgot to specify. So what vertical height does the block reach above its starting point? for part a So I drew the FBD, and summed the forces in both x and y direction as shown above. Once I did that I got -ukN - mgsintheta = ma, so Normal fore is mgcostheta...

A 2.0kg wood block is launched up a wooden ramp that is inclined at a 32 degree angle Initial speed of the block is 8 m/s and the coefficient of kinetic friction is 0.2. a) what height does the block reach above its starting point? b) What speed does it have when it slides back down to...