Recent content by bob tran

Okay, hopefully, I got it this time. \text{At the equilibrium position, PE=0, so:}\\ E=\dfrac{1}{2}mv^2=\dfrac{1}{2}(.5)(1.5)^2=.5625 \text{ J}\\ \text{So, because } \dfrac{E}{2}=KE,\\ E=PE+\dfrac{E}{2}\\ .5625=\dfrac{1}{2}kx^2+\dfrac{.5625}{2}\\ \sqrt{\dfrac{.5625}{k}}=x\\...

Sorry, it is 0.24 m. I just accidentally typed 2.4 m (fixed it in previous post now). But, I am trying to get 0.17 m for part B, not 0.24 m.

k=20, m=0.5, v=1.5, x=?\\ E=PE+KE\\ E=\dfrac{E}{2}+KE\\ E-\dfrac{E}{2}=KE\\ \dfrac{E}{2}=KE\\ \dfrac{E}{2}=\dfrac{1}{2}mv^2\\ E=mv^2\\ \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2=mv^2\\ \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\ ...\\ \texttt{which leads to}\\ x=0.24 \ \texttt{m} Where am I going wrong? Or if I...

Still not sure how I use this for the problem. :frown: \dfrac{E}{2}=PE? But, what exactly am I supposed to do with that? When I substitute this into the equation, it still get 0.24 m as the answer. It seems like I don't understand something fundamental about this or I keep substituting in the...

Still don't see it. :headbang: The only way I got an answer that rounds to 0.17 m was by doing: E=PE+KE\\ E=\dfrac{1}{2}mv^2+\dfrac{1}{2}kA^2=1.125 \ \texttt{J} And then using that value here: \dfrac{E}{2}=PE+\dfrac{KE}{2}\\ E=2PE+KE\\ E=2(\dfrac{1}{2}kx^2)+\dfrac{1}{2}mv^2\\...

Isn't it still 0.24 m? E = PE + KE\\ E = \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\ 2PE = \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\ 2(\dfrac{1}{2}kA^2)= \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\ kA^2= \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\ kA^2-\dfrac{1}{2}mv^2= \dfrac{1}{2}kx^2\\...

\dfrac{PE}{KE}=1 \ ?\\ E = 2PE = 2KE \ ?

Hmm.. I still don't understand how this would get 0.17 m as an answer for part B.

Thanks! v=\dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = 1.5 \dfrac{\texttt{m}}{\texttt{s}}\\ A=\sqrt{\dfrac{(4)(1.5)^2}{100}}=0.300 \ \texttt{m}

Homework Statement A 0.50-kg object is attached to an ideal massless spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. (a) What is the amplitude of vibration? (b) At...

Homework Statement A 2.00-kg object is attached to an ideal massless horizontal spring of spring constant 100.0 N/m and is at rest on a frictionless horizontal table. The spring is aligned along the x-axis and is fixed to a peg in the table. Suddenly this mass is struck by another 2.00-kg...

So would it be like this? W_f=mgd\sin{\theta}+\frac{1}{2}m(v^2_f-v^2_i)-Pd\cos{\theta}

W=-mgh\\ W=-mg(dcos\theta)\\ W=-(700)(9.8)(3\sin{30})\\ W=-10290 \ \texttt{J or } -10300 \ \texttt{J} Wow. I guess I was thrown off because they mentioned P. Thanks! Out of curiosity, how would I find the work that friction does?

Homework Statement In the figure, a 700-kg crate is on a rough surface inclined at 30°. A constant external force P = 5600 N is applied horizontally to the crate. As the force pushes the crate a distance of 3.00 m up the incline, the speed changes from 1.40 m/s to 2.50 m/s. How much work does...

Not sure. It was from a practice worksheet.