A 60 kg block slides along the top of a 100 kg block. The lighter block has an acceleration of 3.4 m/s2 when a horizontal force F= 370 N is applied. Assuming there is no friction between the bottom 100 kg block and the horizontal frictionless surface but there is friction between the blocks. Find the acceleration of the 100 kg block during the time the 60 kg block remains in contact.

I'm at a stand still with this problem.

f = m a

to find the frictional force (ff) between the blocks ... F - ff = m a

370 - ff = 60 * 3.4

the acceleration of the larger block is its mass, divided by the frictional force

I got 0.602 but the answer is 1.66. Any idea why?

To solve this problem, we can start by analyzing the forces acting on each block.

For the lighter 60 kg block:
1. Identify the forces: The only horizontal force acting on the 60 kg block is the applied force F = 370 N.
2. Apply Newton's second law: The net force experienced by the block is equal to the mass of the block multiplied by its acceleration. So, we have F - f = m1 * a, where f is the frictional force between the two blocks, m1 is the mass of the 60 kg block, and a is its acceleration.
3. Determine the unknowns: We need to find the frictional force f.

For the heavier 100 kg block:
1. Identify the forces: The only force acting on the 100 kg block is the frictional force f.
2. Apply Newton's second law: The net force experienced by the block is equal to the mass of the block multiplied by its acceleration. So, we have f = m2 * a2, where m2 is the mass of the 100 kg block, and a2 is its acceleration.
3. Determine the unknowns: We need to find the acceleration a2.

Now, let's solve the equations simultaneously:

For the 60 kg block: F - f = m1 * a
For the 100 kg block: f = m2 * a2

We know the mass of the 60 kg block is 60 kg, the mass of the 100 kg block is 100 kg, and the applied force F is 370 N.

From the information given, the acceleration of the 60 kg block is 3.4 m/s^2. So, we can substitute the known values into the equation for the 60 kg block:

370 N - f = 60 kg * 3.4 m/s^2

Now, we need to determine the frictional force f. To do that, we need to find the normal force between the two blocks.

The normal force is equal to the weight of the upper block, which is given by m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

The normal force is therefore 60 kg * 9.8 m/s^2 = 588 N.

Now, we can calculate the frictional force f using the equation: f = u * N, where u is the coefficient of friction between the two blocks.

Since the surface between the blocks is assumed to be frictional, we can assume a coefficient of friction (u) between the blocks. If there is no specific value given, a reasonable assumption is usually between 0.1 and 0.5. Let's assume a coefficient of friction of 0.3.

Therefore, f = 0.3 * 588 N = 176.4 N.

Now, we can substitute this value of f into the equation for the 60 kg block to solve for the acceleration a:

370 N - 176.4 N = 60 kg * a
193.6 N = 60 kg * a
a = 193.6 N / 60 kg
a ≈ 3.227 m/s^2

So, the acceleration of the 100 kg block during the time the 60 kg block remains in contact is approximately 3.227 m/s^2.