Question

two masses m1 = 5.90 kg and m2 which has a mass 75.0% that of m1, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. If m1 and m2 start from rest, after they have each traveled a distance h = 2.40 m, use energy content to determine the following.
the speed (in m/s) v of the masses
the magnitude (in N) of the tension T in the cord

Answer 1

Given m1 = 5.90 kg, m2 = 0.75 m1 = 4.425 kg, h = 2.40 m, g = 9.80 m/s^2.

Using energy conservation: loss of gravitational potential = gain in kinetic energy
(m1 - m2) g h = 1/2 (m1 + m2) v^2
=> v = sqrt[ 2(m1 - m2) g h / (m1 + m2) ]

v = sqrt[ 2(5.90 − 4.425)(9.80)(2.40) / (5.90 + 4.425) ]
= sqrt(6.72) = 2.59 m/s

Acceleration a = v^2/(2h) = 6.72/(4.80) = 1.40 m/s^2.

Tension (use T = m1(g − a) = m2(g + a)):
T = 5.90(9.80 − 1.40) = 49.6 N

Answers:
- Speed v = 2.59 m/s
- Tension T = 49.6 N