Will the Block Move and What is Its Acceleration?

[vntraderus]

A block whose weight is 35.0 N rests on a horizontal table. A horizontal force of 41 N is applied to the block. The coefficients of static and kinetic friction are 0.650 and 0.420 N, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration?

m/s^2= ?

 

[cristo]

A block whose weight is 35.0 N rests on a horizontal table. A horizontal force of 41 N is applied to the block. The coefficients of static and kinetic friction are 0.650 and 0.420 N, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration?

m/s^2= ?

Do you want someone here to replace the question mark with an answer, or do you want to know what m/s² means?

I should point out that PF rules state that we must see some work from you, and we will not give out final solutions to homework problems.

So, what work have you done so far?

 

[m2003]

Based on the information provided, the block will not move under the influence of the 41 N force. This is because the force of friction, which is equal to the coefficient of static friction (0.650) multiplied by the weight of the block (35.0 N), is greater than the applied force of 41 N. In order for the block to move, the applied force would need to be greater than the force of friction.

If the applied force was increased to be greater than the force of friction, the block would begin to move and the coefficient of kinetic friction (0.420) would come into play. The block's acceleration can be calculated using Newton's second law, which states that the net force applied to an object is equal to its mass multiplied by its acceleration (F=ma). In this case, the net force would be the difference between the applied force and the force of friction. Therefore, the block's acceleration would be (41 N - 22.75 N)/35.0 kg = 0.548 m/s^2. This means that the block would have an acceleration of 0.548 m/s^2 in the direction of the applied force.