Recent content by jjd101

EDIT: I think .86m is wrong, i think i jus ttake 3.36m/s and multiply by .51s and get 1.71m?

so using 1/2kx^2 i get (1/2)(56Nm)(.32m)^2 = 2.87 which through (1/2)mv^2 yields a velocity of 3.79 m/s. Then the decleration is equal to μg = (.52)(-9.8) = -5.096 m/s/s. thru motion equation i got final velocity = 3.36m/s. then thru another motion equation i got time = .51sec for the...

Homework Statement A 400 gram block is launched across a horizontal frictionless table by a horizontal spring that has been compressed 32 cm and has a spring constant of 56 N/m. After going a short distance, the block slides across a 30 cm strip at the edge of the table that has a coefficient...

Thank you!

Thank you!

i normally do include units it's just tedious on the computer haha, and 330 is the product of the mass of the block at 1.5m/s therefore it is a quantity of the momentum i believe? and i believe my mistake is when i do KE=1/2mv^2 i use grams instead of kilograms, therefore it should be...

clockwise = counter clockwise clockwise = 4704+686*9 = 10875 counterclockwise = 490*2 + Fof2ndSupport*8 solving for this 2nd support is a force of 1237N therefore the force of the first support is 2352-1237 = 1115N ?

ohhhhhh, i was forgetting to change grams to kg? so with units as .82 kg now i get T = 1.213sec?

clockwise torque of the board is 1176* 4m = 4704?

220(1.5) = 330 820(v) = 330 v = .4024 KE = 1/2mv^2 KE = 66.4? 2.46 in m

so counter clockwise from first support, 490* 2 counter clockwise from second support 490* 10 clockwise from first support 686*9 clockwise from second support 686*1 these are not even though

ok thank you

force of student one = 9.8*50 = 490 force of student 2 = 9.8*70 = 686 force of beam = 9.8*120= 1176 Total down force =2352 = Support 1 force + support 2 force?

sorry i meant power, 540000 is the value for power, isn't omega 1500 rpm converted to rad/sec which is 157.1rad/sec?

T= 2(pie)(sqroot(m/k)) T= 38.4sec?