(a) Since m1 > m2, m1 will fall freely while m2 will move upward. The only force acting on both blocks is gravity, so we'll use the following equation for motion under constant acceleration:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (0 in this case), a is the acceleration, and s is the distance traveled.
For m1, we can write:
v1^2 = 0^2 + 2a1h
where v1 is the final velocity of m1, a1 is the acceleration of m1, and h is the distance m1 falls.
For m2, we can write:
v2^2 = 0^2 + 2a2h
where v2 is the final velocity of m2 and a2 is the acceleration of m2.
Since the string connecting the blocks is inextensible, both blocks must have the same magnitude of acceleration, so we can write a1 = a2 = a.
The net force acting on m1 is m1g - T, where T is the tension in the string, and that on m2 is T - m2g. By applying Newton's second law (F = ma), we get:
m1g - T = m1a (1)
T - m2g = m2a (2)
Adding equations (1) and (2), we get:
m1g - m2g = m1a + m2a
a(m1 + m2) = g(m1 - m2)
a = g(m1 - m2) / (m1 + m2)
Now we can substitute this value of a in the equation for v1^2:
v1^2 = 2 * (g(m1 - m2) / (m1 + m2)) * h
v1 = sqrt[2 * (g(m1 - m2) / (m1 + m2)) * h]
This is the expression for the speed of m1 just as it reaches the floor.
(b) Now substituting the given values of m1, m2, and h:
m1 = 6.1 kg
m2 = 4.5 kg
h = 3.1 m
g = 9.81 m/s² (approximately)
v1 = sqrt[2 * (9.81(6.1 - 4.5) / (6.1 + 4.5)) * 3.1]
v1 = sqrt(2 * (9.81 * 1.6 / 10.6) * 3.1)
v1 ≈ 3.16 m/s
So, the speed of m1 just as it reaches the floor is approximately 3.16 m/s.
(c) To find the speed of each block when m1 has fallen a distance of 1.2 m, we can simply replace h with 1.2 m in the expression for v1:
v1 = sqrt[2 * (g(m1 - m2) / (m1 + m2)) * 1.2]
v1 = sqrt(2 * (9.81 * 1.6 / 10.6) * 1.2)
v1 ≈ 1.94 m/s
Since the blocks have the same magnitude of acceleration, their speeds will be equal at any given instant. Therefore, the speed of m2 will also be approximately 1.94 m/s when m1 has fallen a distance of 1.2 m.
Two blocks are connected by a light string that passes over a frictionless pulley as in the figure below. The system is released from rest while m2 is on the floor and m1 is a distance h above the floor.
(a) Assuming m1 > m2, find an expression for the speed of m1 just as it reaches the floor. (Use any variable or symbol stated above along with the following as necessary: g.)
(b) Taking m1 = 6.1 kg, m2 = 4.5 kg, and h = 3.1 m, evaluate your answer to part (a).
(c) Find the speed of each block when m1 has fallen a distance of 1.2 m.
1 answer
1 answer