An atwood machine consists of a massless string connecting two masses over a massless, frictionless pully. In this case, the masses are 1.70kg and 1.65kg. The system is released from rest with the 1.7kg mass 2.15m above the floor and the 1.65kg mass on the floor. (a) what is the acceleration of the 1.7kg mass? of the 1.65kg mass? (b) what is the speed of the 1.7kg mass just before it hits the floor? (c) how long does it take the 1.7kg mass to reach the floor?
The acceleration of a frictionless, massless Atwood machine is
a = (M1-M2)g/(M1+M2)
Se if you can use that information to do the rest of the problem
To solve this problem, we can use Newton's second law of motion and the principles of conservation of energy.
(a) To find the acceleration of each mass, we'll start by analyzing the forces acting on them. The net force on each mass can be given by the difference in tension forces on either side of the pulley.
Let's assume that the downward direction is positive. The gravitational force acting on the 1.70 kg mass is:
F1 = m1 * g
where m1 = mass of the 1.70 kg mass and g = acceleration due to gravity (9.8 m/s^2).
The gravitational force acting on the 1.65 kg mass is:
F2 = m2 * g
where m2 = mass of the 1.65 kg mass.
Since the masses are connected by a string, the tension in the string on the side of the 1.70 kg mass (T1) is greater than the tension on the side of the 1.65 kg mass (T2). So, we can write:
T1 = m1 * g + m1 * a
T2 = m2 * g - m2 * a
where a = acceleration of the system.
The net force on the 1.70 kg mass is given by:
Fnet1 = T2 - T1 = m2 * g - m2 * a - (m1 * g + m1 * a) = (m2 - m1) * g - (m1 + m2) * a
Setting Fnet1 equal to m1 * a (according to Newton's second law), we get:
m1 * a = (m2 - m1) * g - (m1 + m2) * a
Simplifying the equation, we can solve for a:
a = (m2 - m1) * g / (m1 + m2)
Plugging in the given values, we can find the acceleration of the 1.70 kg mass.
(b) To find the speed of the 1.70 kg mass just before it hits the floor, we can use the principle of conservation of mechanical energy. The potential energy converted into kinetic energy when the mass drops, and there is no energy loss due to friction.
The initial potential energy (PEi) of the 1.70 kg mass is given by:
PEi = m1 * g * h
where h is the height (2.15 m) above the floor from which the mass is released.
The final kinetic energy (KEf) just before hitting the floor is given by:
KEf = (1/2) * m1 * v^2
where v is the speed of the 1.70 kg mass just before hitting the floor.
Since energy is conserved, we have:
PEi = KEf
Simplifying the equation, we can solve for v:
v = sqrt((2 * m1 * g * h) / m1)
Plugging in the given values, we can find the speed of the 1.70 kg mass just before it hits the floor.
(c) To find the time it takes for the 1.70 kg mass to reach the floor, we can use the equation of motion:
h = (1/2) * g * t^2
Simplifying the equation, we can solve for t:
t = sqrt((2 * h) / g)
Plugging in the given values, we can find the time it takes for the 1.70 kg mass to reach the floor.