Recent content by aqryus

definitely an 8 on the worksheet

First I found work: W=(3.85x10^5)(2.45x10^8) W= 9.43x10^13 Then used that for difference of kinetic energy: 9.43x10^13 = (1/2) (4.55x10^4)v2^2 - (1/2)(4.55x10^4)(1.22x10^4)^2 9.43x10^13 = (22750)v2^2 - 3.386x10^12 9.43x10^13 + 3.386x10^12 = (22750)v2^2 9.77x10^13 = 22750v2^2 9.77x10^13/22750...

I'm a little confused because my teacher used Bill's 500J of work for the kinetic energy equation and I don't understand why. I used the net work, so 300J, to find the speed and I'm not sure why that's wrong. Wouldn't friction make the wagon move slower than if there was no friction? So why...

i used t=d/v for the x direction using the formula t=250/vcos55 and subbed that into the formula for the y direction d = (v) (t) + (0.5) (a) (t)^2 35 = (vsin55) (250/vcos55) - (4.9) (250/vcos55)^2 canceled out the first two v 35 = sin55 (250/cos55) - (306250/v^2cos3025) 35 = 357 -...

Okay so after the eraser is shot the applied force is zero and only friction force does work on it right?

The answer is .32m. I set the elastic potential energy as equal to the work, but at first I put the force in the work equation as (F elastic - F kinetic friction) times distance and rearranged. 1/2kx^2 = (kx-Ff) d (0.5) (22) (0.035)^2 = (22 x 0.035-0.042) d 0.013475= 0.728 d 0.013475/0.728 = d...