Question

Two identical blocks 3.0 kg are stacked on top of each other. The bottom block is free to slide on a frictionless surface. The coefficient of static friction between the blocks is 0.35.What is the maximum horizontal force that can be applied to the lower block without the upper block slipping?

Answer 1

The maximum horizontal force that can be applied to the lower block without the upper block slipping is 10.29 N.

Step-by-step explanation:

To determine the maximum horizontal force that can be applied to the lower block without the upper block slipping, we need to consider the friction between the blocks. The formula for static friction is F_s = u_s * N, where F_s is the maximum static frictional force, u_s is the coefficient of static friction, and N is the normal force between the blocks. In this case, the normal force is equal to the weight of the upper block, which is 3.0 kg * 9.8 m/s^2 = 29.4 N. Hence, the maximum static frictional force is 0.35 * 29.4 N = 10.29 N.

Answer 2

The maximum horizontal force that can be applied to the lower block without the upper block slipping is 10.29 N, calculated using the static coefficient of friction multiplied by the normal force exerted by the weight of the upper block.

Step-by-step explanation:

The maximum horizontal force that can be applied to the lower block without the upper block slipping can be determined using the static coefficient of friction between the blocks and the normal force exerted by the upper block. Since the blocks are identical and each has a mass of 3.0 kg, the weight of the upper block (W = mg) provides the normal force (N = W), which is 3.0 kg × 9.8 m/s² = 29.4 N.

We can then calculate the maximum force of static friction (fₙ) that can act on the upper block using fₙ = μₙ N. With a coefficient of static friction (μₙ) of 0.35, the maximum static frictional force is fₙ = 0.35 × 29.4 N = 10.29 N.

This is the maximum frictional force that can act between the two blocks before the upper block slips. Therefore, the maximum horizontal force applied to the lower block that will not cause the upper block to slip is 10.29 N.