Will a block on a table move with a 36N force and what will its acceleration be?

[explorer]

A block whose weight is 45N rests on a horizontal table.
A horizontal force of 36N is applied to the block .
The coefficients of static and kinetic friction are 0.65 and 0.42, respectively.
Will the block move under the influence of the force, and if so, what will be the block's acceleration?

Thanks for you help.

 

[VBPhysics]

Explorer, this sounds a lot like a homework problem to me. There is a place on the board for this, and I'll request it be moved. To help you with your problem I'll give you some general ideas about this problem.

1st you have the normal force and both coeficients of friction. Look up the equation for how coeffiecients and normal force work together to produce a resistive force.

2nd Calculate what force must be overcome to get the block moving(static friction). Do you have enough force being applying to overcome this barrier?(that's part one)

3rd If the block does slide, calculate the net force on the system( the force of you pushing mius kinetic friction)?

4th use the net force on the system & the known mass of the system with Newtons second law(F=ma) to determine acceleration.

Hope this helps

 

[M98Ranger]

Based on the given information, it is possible for the block to move with a 36N force if the force is greater than the maximum static friction force of 29.25N (45N x 0.65). If the applied force is greater than the maximum static friction force, the block will begin to move with an acceleration of 0.8 m/s^2 (36N / 45kg).

Once the block is in motion, the kinetic friction force will come into play and act in the opposite direction of the applied force. The acceleration of the block will then depend on the net force acting on it, which will be the difference between the applied force and the kinetic friction force.

To calculate the kinetic friction force, we can use the formula Fk = μkN, where μk is the coefficient of kinetic friction and N is the normal force (equal to the weight of the block, 45N). This gives us a kinetic friction force of 18.9N (0.42 x 45N).

Therefore, the net force acting on the block will be 17.1N (36N - 18.9N), resulting in an acceleration of 0.38 m/s^2 (17.1N / 45kg).

In summary, the block will move with an acceleration of 0.8 m/s^2 if the applied force is greater than 29.25N, and with an acceleration of 0.38 m/s^2 once it is in motion.