Three blocks with masses m1 = 7.8 kg, m2 = 14 kg, and m3 = 19 kg, respectively, are attached by strings over frictionless pulleys. The horizontal surface exerts a 32 N force of friction on m2. If the system is released from rest, use energy concepts to find the speed (in m/s) of m3 after it moves down 3.0 m.
The picture shows m1, m2, and m3 are all connected by a string. m1 is hanging off the table by the string on the left, m2 is sitting on the table and is experiencing friciton, and m3 is haning off the table by the string on the right. Please show your work, I can't seem to figure this out!
- Favorite Answer
Through conservation of energy
h((m3-m1)*g-32) = (m1+m2+m3)/2*V^2
V = √((3*9,806*(19-7.8)-32*3)/((19+7.8+14)/2) = 3.4 m/sec^2
accelerating force AF = (m3-m1)*g-32
acceleration a :
a = AF / (m1+m2+m3) = ((19-7.8)*9.806-32)/(7,8+14+19) = 1.9 m/sec^2
V = √ 2*a*h = √ 2*3*1.91 = 3.4 m/sec
- NCSLv 72 months ago
Some of the initial PE of the arrangement gets converted into KE and friction work:
(19 - 7.8)kg * 9.8m/s² * 3.0m = ½*(7.8+14+19)kg*V² + 32N*3.0m
solves to V = 3.4 m/s
If instead you were to use a kinematics approach, you'd observe that the net motive force is
F = (19-7.8)kg*9.8m/s² - 32N = 78 N
and so the system acceleration is
a = F / Σm = 78N / 40.8kg = 1.9 m/s²
and then V = √(2ad) = √(2 * 1.9m/s² * 3.0m) = 3.4 m/s
Hope this helps!
We need the picture to know how the blocks are attached.