Question

Q=125

m/s Problem 2, A mine car has a mass m = 250 kg and it is hoisted up the incline using the cable and motor M. The force in the cable is F(t) = 1050 + (30) t?, N where t is in seconds. If the car has an initial velocity v. = 3 m/s at s = 0 and t = 0. Q4) What is the velocity of the car when t = 3 s, in (m/s). (5) What is the distance it moves up the plane when t = 3 s, in (m).

Answer 1

Given that Q = 125 î Î Î mass of the Cant m= 250 kg - The force in the cable is F(t) = 1050+ (30) (62) If t=0 - S=0 & V, =3m/s. - tension in the wire. FCE) = 1050 + (3)(125) (62). = 1050+ 375 (t²) -> N (250) (9.81) = 2452.5 N N it FCE) o Wsingol (ma) Wloszo' 15 O = 28.07 W=21525 N tand=8 from the eavilibrium Condition 15 8 = 28.07 Efn=0 os (wsingo) + (ma) - FCE). 28-07 (2452.5) (sin 28.07) + (250) (a) = F(t)

154.ll + 250 (a) = 1050 +375 (t²). = ) 250 (a) = -104.02 +375 (2) a a= -0.416 + 1.5 (E²). 1 we know that du dv de dv dt 0.416 & 1.5(6²) =) Sav = (-ouro -OouI6 + 1.56²) at 2) -O. UGCE) + 15t3 tci w Boundary Condition o when tzo - Vesmls. 01=3m/s V=-0.016 (L) + 0.5 t) +3 A when t=3sec ) Velocity of mine Cart. V=-0.416 (3-0) + 0.5 (336) +3. zasse 223se 15.625 mis

from @ =) V = -0.016 CE) +0.5(+3)+3 dt -) e da = -0.016 (t) +0,5 (+3)+3 Sds = - Jf-0.416(b) + 05 (69)+3 ]dt S= -0.11642 ) + 0.5(+36 +C, S = -0.208 (€²) + 0.125 (4") +3E+G -) when tzo > S=0 -) C2=0. i S = -0.208 (62) + 0.125 CE")t3t sol when t=3sec Skesses = -0.208 (3) + 0.125 (3)4+3)(3) s = 17.253 m