HW6 #1

Does anyone know how to solve #1? I have already tried change in energy over time, P=Fv, etc...

HW6 #17

Hey I am having a problem getting #17 to work out on the homework with my numbers. When I tried solving it with other people's numbers the same way, I got the right answer, but mine arent working out. I have an 18 kg child on a playground swing moving with a speed of 4.7 m/s at the lowest point, and the swing is 1 meter long. I need to find the angle the string makes with the vertical. First I tried using conservation of energy with K=U, so .5mv^2=mgh. The m's cancel out and I am trying to solve for h, height above the lowest point. I end up with .5(4.7^2)=9.81h, and when I solve for h, I get 1.12, which doesn't work because it is higher than the length of the string. I also tried solving it by setting cos(theta)=(centripital acceleration)/g, and the acceleration is greater than g so it doesnt work. Please let me know if I am doing something wrong! Thanks!

HW5 #18

Hey everyone, I just wanted to share a helpful tip for #18. Be careful when converting cubic kilometers to cubic meters. It's not a 1:1,000 ratio because the value is cubed.

HW5 #14

For what length of time should the power be calculated? Is this question asking for the power generated per second/hour/day?

HW4 #5

Can somebody please help me with this question? I was able to answer the three preceding steps, but I seem to be stuck on this one. I attempted to calculate the horizontal components of T1 and T2 using trigonometry, but this yielded an incorrect solution. I then tried adding these values together, but this was wrong too. Please help!

HW4 #17

Can anyone give me some help on how to do 17? thanks

HW4 #15

If you are having trouble with this problem but you managed to do the other force problems, you might want to take another look at your force diagrams (free body diagrams).

The smaller block hangs over the edge of a table (over a pulley), and is attached to a string. There is a tension T upward, a force due to gravity (Fg = m*g) downward, and the block's resulting acceleration is downward. You use Newton's 2nd law to get an expression for the tension: ∑F=Fg-T=ma so, rearranging and writing Fg=mg, T=mg-ma.

Now look at the larger block. It has a force of gravity downward, a normal force upward, and 2 ropes pulling to the right. That makes it T to the right and another T to the right, so there is 2T pulling to the right (look at the pulley: it makes the string pull twice). With "right" as the positive direction, Newton's 2nd law becomes ∑F=2T=ma. Plug in the expression you had above for T and solve for the acceleration. "The rest is just algebra."

Hope that helps. Good luck.