One on an Incline--One Hanging A block of mass mA = 4.20 kg on a frictionless inclined plane of angle 30.0° is connected by a cord over a massless, frictionless pulley to a second block of mass mB = 2.20 kg hanging vertically.(a) What is the magnitude of the acceleration of each block? m/s2 (mA) m/s2 (mB)(b) What is the direction of the acceleration of the hanging block?not enough informationmB accelerates downwards mB accelerates upwards(c) What is the tension in the cord? N
Question
One on an Incline--One Hanging A block of mass mA = 4.20 kg on a frictionless inclined plane of angle 30.0° is connected by a cord over a massless, frictionless pulley to a second block of mass mB = 2.20 kg hanging vertically.(a) What is the magnitude of the acceleration of each block? m/s2 (mA) m/s2 (mB)(b) What is the direction of the acceleration of the hanging block?not enough informationmB accelerates downwards mB accelerates upwards(c) What is the tension in the cord? N
Solution
To solve this problem, we need to use Newton's second law (F = ma) and the equations for the forces acting on each block.
(a) The forces acting on block A are its weight (mAg), the normal force (nA), and the tension in the cord (T). The forces acting on block B are its weight (mBg) and the tension in the cord (T).
For block A, the force along the incline is mAgsin(30) - T. According to Newton's second law, this is equal to mA*a, where a is the acceleration of the block. So we have:
mAgsin(30) - T = mA*a ----(1)
For block B, the force is T - mBg. According to Newton's second law, this is equal to mBa. So we have:
T - mBg = mBa ----(2)
We can solve these two equations simultaneously to find a and T. Adding the two equations gives:
mAgsin(30) - mB*g = (mA + mB)*a
Solving for a gives:
a = (mAgsin(30) - mB*g) / (mA + mB)
Substituting the given values gives:
a = (4.20 kg * 9.8 m/s^2 * sin(30) - 2.20 kg * 9.8 m/s^2) / (4.20 kg + 2.20 kg) = 1.96 m/s^2
So the magnitude of the acceleration of each block is 1.96 m/s^2.
(b) The direction of the acceleration of the hanging block is downwards. This is because the tension in the cord is less than the weight of the block, so the net force is downwards.
(c) To find the tension in the cord, we can substitute the value of a into equation (2):
T = mBa + mBg = 2.20 kg * 1.96 m/s^2 + 2.20 kg * 9.8 m/s^2 = 25.84 N
So the tension in the cord is 25.84 N.
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