omega = omega_o + alpha t theta = theta_o + omega_o t + 1/2 alpha t^2 theta = th
ID: 1557938 • Letter: O
Question
omega = omega_o + alpha t theta = theta_o + omega_o t + 1/2 alpha t^2 theta = theta_o + omega t - 1/2 alpha t^2 omega^2 = omega_o^2 + 2 alpha (theta - theta_o) p = mv P = W/t v_1 - v_2 = -(v' _1 - v' _2) f = mu N W = Fd W = delta K K = 1/2 mv^2 U = mgh E = K + U A wheel rotates 4500 times before reaching a top speed of 34.0 rev/s. If its angular acceleration is 11.0 rad.s^2 how long did it rotate to reach its speed? An object of m_1 = 2.0 kg moving with a velocity -12.0 m/s collides head-on with a moving object whose mass m_2 = 6.0 kg. Given that the collision is elastic, and the velocity of the second mass after the collision -2.5 m/s what is the initial velocity of m_2 and final velocity of mi? Neglect friction. A box of mass 5.0 kg starts from rest and slides down an incline plane of angle 35 degree which has a base of 0.7 m. Half of the path down the incline is frictionless; the other half has a coefficient of 0.07. Calculate: a. The energy lost going down the incline in joules. b. The speed of the box just as it reaches the ground in m/s. A 15.0 kg object is accelerated as it slides across a surface where its velocity changes 8.0 m/s to 35.0 m/s over the course of 2.5 seconds. How much power was required?Explanation / Answer
1) w = w0+alpha*t
w = alpha*t
t = 34*2*31.4/11 = 19.41 sec
2) v1 = 2m2u2+(m1-m2)u1/(m1+m2)
v2 = 2m1u1+(m2-m1)u2/(m1+m2)
(m2-m1)*u2 = -20+48 = 28
u2 = 7 m/s
v1 = 2*6*7+12*4/8 = 16.5 m/s
3) l = 0.7/cos35 = 0.854 m
for first half w1 = 5*9.8*sin35*0.854/2 = 12 J
for second half w2 = -f*s = -mue*mgcos35*0.854/2 = -1.199 J
w tot = 10.801 J
w = k2-k1
k1 = 0
v = sqrt(2w/m) = 2.08 m/s
4) w = 1/2m(v2^2-v1^2) = 8707.5 J
power = w/t = 3483 W
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