Question

Two masses M1 2 kg and M2-4 kg are attached by a string as shown. They start from rest and move with no friction until they reach a velocity of 6.5 m/s. When do they reach that speed, in s? Part A 0.68 O 4.3 O 2.0 O1.4 O 3.3 Submit My Answers Give Up

Answer 1

given mass M1 = 2 kg
M2 = 4 kg
attached to vertical pulley

a. initial velocity u = 0 m/s
final velocity, v = 6.5 m/s

let acceleration of the system be a
then from newtons second law
(M2 - M1)g/(M2 + M1) = a
a = 3.27 m/s/s

so time taken to reach final velocity = t
then v = at
t = 1.987 s

so the answer is t = 2 s

A. mass of elevator,. M = 1000 kg
   counter weight mass, m = 900 kg

   so when the system begins to fall,
   acceleration of the system, a = (M - m)g/(M + m)= 0.5163 m/s/s

   so net force on the system = (M + m)a = 981 N ( downwards)
   so the answer is 980 N

A. tilt angle = theta
then given tan(theta) = 0.02/1
theta = 1.145 deg

so mass of glider = m = 1 kg
force parallel to the beam on glider = mgsin(theta) = 0.196 N
hence the answer is 0.2 N