Two blocks are connected by a string and pulled by a Force F = 50N. If the coefficient of friction between the blocks and the ground is 0.2, friction force between the block B and the ground is:

When two blocks are connected by a string and pulled with a force F = 50N, we analyze the friction acting on the blocks. The coefficient of friction between the blocks and the ground is given as μ = 0.2. Let us calculate the frictional forces step by step:

Step 1: Friction Force Between Block A and the Ground

Using the formula for maximum static friction, f_s = μ × N, where N is the normal force acting on block A:

f_s = 0.2 × 300N = 60N

The frictional force between block A and the ground is 60N. This value exceeds the applied force F = 50N. As a result, the blocks do not move.

Step 2: Implication for Tension in the String

Since the applied force F is less than the maximum static friction force, the blocks remain stationary. The tension in the string connecting the two blocks is zero, as there is no relative motion between the blocks. Hence:

T = 0

Step 3: Friction Force Between Block B and the Ground

Since the two blocks are connected by a string and there is no movement, the friction force acting on block B is also zero. The applied force is insufficient to overcome the frictional resistance of block A, so block B remains stationary as well.

f_B = 0