Solve Incline Problem: Find Highest Point Reached by Block

  • Thread starter Thread starter milkyway11
  • Start date Start date
  • Tags Tags
    Incline
milkyway11
Messages
15
Reaction score
0
question says: a block is given an initial velocity of 2m/s up a frictionless plane inclined at 60 degree to the horizontal. what is the highest point reached by the block?

I tried to figure out how to solve the problem by thinking in order to find the highest point, i need the distance it traveled then I can use the sin to find the height. However, in order to find distance traveled, I will have to find the acceleration up the incline and this is where I get stuck. I tried to use kinematics to find the acceleration knowing final velocity will be 0, but I don't know the time spent traveling so I can't solve for acceleration!

Please help me solve it!
 
Physics news on Phys.org
Use conservation of energy.

At the top, the kinetic energy is 0 so it is all potential energy. At the bottom, it is all kinetic (if you set V=0 at that point). The two values are equal...hence 1/2 mv^2 at bottom = mgh at top, but that h is the vertical height, so you need to use trigonometry to find the distance up the incline...
 
Of course, the acceleration (if you do not know the conservation of energy theorem) is easy enough to find. The only force acting down the block is the accelaration due to gravity (taking into account trigonometry to find it's component down the incline...)
 
thank you, this helps a lot!

Berko said:
Use conservation of energy.

At the top, the kinetic energy is 0 so it is all potential energy. At the bottom, it is all kinetic (if you set V=0 at that point). The two values are equal...hence 1/2 mv^2 at bottom = mgh at top, but that h is the vertical height, so you need to use trigonometry to find the distance up the incline...
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top