A 12.0 kg block sits on a frictionless table. On top of this block is placed a 5.00 kg block. The coefficient of static friction between the two blocks is 0.70. a) What is the maximum force that can be applied horizontally to the 5.00 kg block before it slips?
b) What is the acceleration of the
5.0 kg block (just before it would slip)?
b) What is the maximum force that can be applied horizontally to the 12.0 kg block before the 5.0 kg block begins to slip?
To find the maximum force that can be applied horizontally to the 5.00 kg block before it slips, we need to consider the static friction between the two blocks.
First, let's calculate the maximum static friction force between the blocks. The formula for static friction is F_static = coefficient * normal force, where the normal force is the force pressing the two blocks together.
The normal force can be found by calculating the weight of the upper block, which is the mass multiplied by the acceleration due to gravity (9.8 m/s^2). So, the weight of the upper block is 5.00 kg * 9.8 m/s^2 = 49 N.
Therefore, the normal force pressing the blocks together is equal to the weight of the upper block, which is 49 N.
Now, calculate the maximum static friction force by multiplying the coefficient of static friction (0.70) by the normal force (49 N): F_static = 0.70 * 49 N = 34.3 N.
Hence, the maximum force that can be applied horizontally to the 5.00 kg block before it slips is 34.3 N.
To find the acceleration of the 5.0 kg block just before it slips, we need to use Newton's second law, which states that Force = mass * acceleration (F = m * a).
The force applied to the 5.00 kg block is the maximum force that can be applied before it slips, which is 34.3 N. The mass of the 5.00 kg block is 5.00 kg.
Thus, plugging these values into the formula, we get: 34.3 N = 5.00 kg * a.
Therefore, the acceleration of the 5.0 kg block just before it slips would be 34.3 N / 5.00 kg = 6.86 m/s^2.
Lastly, if we want to find the maximum force that can be applied to the 12.0 kg block before the 5.0 kg block begins to slip, we need to consider the maximum force of static friction between the 12.0 kg block and the table.
The maximum static friction force between the 12.0 kg block and the table can be calculated in the same way as before.
The weight of the 12.0 kg block is 12.0 kg * 9.8 m/s^2 = 117.6 N, which is the normal force.
Then, the maximum static friction force is 0.70 * 117.6 N = 82.32 N.
Therefore, the maximum force that can be applied horizontally to the 12.0 kg block before the 5.0 kg block begins to slip is 82.32 N.