Question

What is the acceleration of M across the frictionless table? Suppose M = 2.4 kg and m = 1.2 kg. Hint: Think carefully about the acceleration constraint. (Figure 1) Express your answer with the appropriate units.

Answer 1

The acceleration of M across the frictionless table is \
`( (g \cdot (m - M))/((M + m)) \)`, where (g ) is the acceleration due to gravity, ( m ) is the mass of the hanging object, and ( M ) is the mass of the block on the table.

Step-by-step explanation:

The acceleration of the system can be determined by applying Newton's second law and considering the tension in the string. The net force acting on the system is the difference between the gravitational force on the hanging mass
`(\(m \cdot g\))` and the tension in the string (T). The equation representing this is:

`\[ (m + M) \cdot a = m \cdot g - T \]`

The tension in the string is also the force acting on the block (M) and is equal to
`\(M \cdot a\)`. Substituting this into the equation, we get:

`\[ (m + M) \cdot a = m \cdot g - M \cdot a \]`

Now, solving for acceleration (a):

`\[ a = (g \cdot (m - M))/((M + m)) \]`

This expression represents the acceleration of the block (M) across the frictionless table. The negative sign in the numerator indicates that the acceleration is in the opposite direction to the gravitational acceleration. The acceleration is influenced by the difference in mass between the hanging object (m) and the block (M), as well as the total mass (M + m).

In summary, the final expression
`\( (g \cdot (m - M))/((M + m)) \)` gives the acceleration of the block on the frictionless table, taking into account the mass of the hanging object and the block itself.

Answer 2

To determine the acceleration of mass M on the frictionless table, Newton's second law is applied to both masses. Solving for tension and substituting into the other equation yields an acceleration of approximately 3.27 m/s² for mass M.

Step-by-step explanation:

The problem in question involves a two-block system with a frictionless table and a frictionless pulley. To find the acceleration of the larger mass M on the table, Newton's second law of motion can be applied to both masses. In this case, the only force acting on the mass M (block on the table) is the tension T in the rope, and the only forces acting on the smaller mass m (hanging block) are tension T and the gravitational force mg, where g is the acceleration due to gravity (9.8 m/s²).

For mass M, the equation is T = Ma, where a is the acceleration of the system. For mass m, we use mg - T = ma. Since the string is inextensible and the pulley is frictionless, the acceleration of both masses must be the same; hence, we can call it 'a' for both. Solving these two equations simultaneously gives us the value of acceleration 'a'.

First, we solve for T from T = Ma and then substitute that into mg - T = ma to find the acceleration.

Starting with T from the first equation, T = 2.4kg * a.
Substituting T into the second equation, we get:
1.2kg * 9.8 m/s² - 2.4kg * a = 1.2kg * a.
This simplifies to:
11.76 N = 3.6kg * a.
Hence, dividing both sides by 3.6kg, we get:
a = 11.76 N / 3.6kg ≈ 3.27 m/s².