I know the answer to this question , I just need to know how to get to the answer!?

2010-06-18 4:57 pm
Block A has a mass of 3.00 kg and rests on a smooth table and is connected to block B, which has a mass of 2.00 kg after passing over an ideal pulley. Block B is released from rest. What is the acceleration of the masses? I know the answer is 3.92 m/s squared.

回答 (4)

2010-06-18 5:10 pm
✔ 最佳答案
F=mass*acceleration
because block be is being pulled down by gravity, it has 2*9.8 or 19.6 newtons of force at work (9.8 is acceleration due to gravity). however, this force is distributed over 5 kg, so you need to divide it by five to get the acceleration. 19.6/5 = 3.92
hope this helps
2010-06-19 12:03 am
The force of gravity acting on the two masses is: 9.8 N/kg * 2.00 kg = 19.6 N
That force is acting on 5 kg of mass (both blocks combined).
So the overall acceleration is: 19.6 / 5 = 3.92
2010-06-19 2:10 am
Let acceleration of masses be a.

F = ma
Gravitational force does not accelerate 3kg mass because 3kg mass is accelerating in the horizontal direction while gravitational force is in the vertical direction.

combined mass * acceleration = gravitational pull on 2kg mass

(3 + 2) * a = 2 * 9.81
a = 2 * 9.81 / (3 + 2)
a = 3.924
2010-06-19 12:16 am
Let the gravitational acceleration, g = 9.8 m/s².

For the hanging block of mass, m.B, the tension in the line is: T = (m.B)g - (m.B)a

For the block on the table of mass, m.A, the tension in the line is: T = (m.A)a

Since the tensions are equal:

(m.A)a = (m.B)g - (m.B)a

Solving for the acceleration, a:

(m.A)a + (m.B)a = (m.B)g

a[(m.A) + (m.B)] = (m.B)g

a = (m.B)g/[(m.A) + (m.B)] = (2.0)(9.8)/(3.0 + 2.0) = 3.92 m/s²


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