A 2 kg block is placed on the surface of a table. The coefficient of static friction between the block and the table is 0.5. If a horizontal force of 2.5 N is applied on the block, then the block will
Correct: B
The force of static friction is given by the equation F = μs*N, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the block, which is 2 kg * 9.8 m/s^2 = 19.6 N. So, the force of static friction is 0.5 * 19.6 N = 9.8 N. Since the applied force is 2.5 N, which is less than the force of static friction, the block will not move