What acceleration would a 175 kg box have on this ramp?
Can you please show me HOW you got the answer too?
THANK YOU in Advance!!
2007-12-09 08:27:21 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
In general
F=Fd-f
ma= mg sin(A) - umg cos(A)
a= g(sin(A) - u cos(A))
Than means that acceleration is independent of mass and the for 75 kg will be the same as for the 175 kg mass. Acceleration depends on two factors; angle A and coefficient of friction u. Since u is the same and angle is the same(25 degrees) then
a= 3.6 m/s^2
2007-12-09 08:34:40 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
you ought to locate the internet tension, back out the friction tension, and divide via the conventional tension. i will practice you. Taking words interior the path of commute Fnet = Wsin(25) - Ff the place W = weight and Ff = friction tension. W = mg and Fnet = ma ma = mg - Ff Ff = = mg - ma = m(g-a) = seventy 5 kg (9.80 one - 3.6) m/s^2 = 465.seventy 5 N you besides might comprehend that interior the path perpendicular to commute, the acceleration is 0, so Fn (commonplace tension) = Wcos(25) = seventy 5 cos25 * 9.80 one = 666.80 two N considering that Ff = u*Fn , 465.seventy 5 N = u* 666.80 two N u = 465.seventy 5/666.80 two = 0.698 <-----this is your friction coefficient.
2016-11-15 01:21:58 · answer #2 · answered by ? 4 · 0⤊ 0⤋