Answer:
1) a = 3.07 m/s²
2) α = 2.05 rad/s²
3) T1 = 28.98 N
4) T2 = 27.05 N
5) x =3.5 meter
6) vf =4.64 m/s
7) ωf = 30.93 rad/s
Explanation:
<u>Step 1:</u> Data given
Mass of box 1 = 4.3 kg
Mass of box 2 = 2.1 kg
Mass of pulley = 2.5 kg
Radius = 0.15m
1) <u>What is the linear acceleration of the left box? </u>
A force F at the pulley rim gives an accelerating torque (T = F * r) and an angular acceleration dω/dt (rad/s²)
T = F * r = I*dω/dt
Fp = (I/r) * dω/dt
⇒ with I = ½mr²
⇒ with dω/dt =a/r
Fp = (½mr²/r) * a/r
Fn = ½ma = (4.2-1.9) * 9.81 = 22.56 N
Fn - Fp = (m1 + m2)a
Fn - ½ma = (m1+m2)a
Fn = a(m1 + m2 + ½m)
a = Fn / (m1 + m2 + ½m)
a = 22.56 N/ (4.2 + 1.9 + ½*2.5)
a = 3.07 m/s²
2) <u>What is the angular acceleration of the pulley? </u>
angular acceleration α = a/r
angular acceleration α = 3.07 /0.15 = 2.05 rad/s²
<u>3) What is the tension in the string between the left mass and the pulley?</u>
T1 = m1*(g-a)
T1 = 4.3*
(9.81-3.07)
T1 = 28.98 N
<u>4) What is the tension in the string between the right mass and the pulley? </u>
T2 = m2(g+a)
T2 = 2.1*(9.81 +3.07)
T2 = 27.05 N
<u>5) The boxes accelerate for a time t = 1.51 s. What distance does each box move in the time 1.51 s?</u>
<u />
Initial speed v0 = 0 m/s
Distance x= v0*t + ½at²
x =0*1.51 + ½* 3.07*1.51²
x =3.5 meter
<u>6) What is the magnitude of the velocity of the boxes after the time 1.51 s?</u>
vf = the final speed after a movement of 3.5 meter
vf²=v0²+2ax
vf² = 2*a*x
vf² = 2*3.07*3.5
vf² = 21.49
vf = √(21.49) = 4.64 m/s
<u>7) What is the magnitude of the final angular speed of the pulley? rad/s</u>
final angular speed ωf = vf/ r
ωf = 4.64 m/s/ 0.15m
ωf = 30.93 rad/s
