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A 7.5-kg block is sliding down a wall with constant velocity. The coefficient of static friction between the block and the wall is ?s = 0.70. The coefficient of kinetic friction between the block and the wall is ?k = 0.40. What is the magnitude of the force (in N) pressing the block against the wall? Use g = 9.79 m/s2. NEVER include units with any answer to a numerical question.

Respuesta :

tiara143
20.56  Newtons

The force appled to appose the static friction is
 
F = [tex] u_{s} [/tex]× m × g 

Where is u_{s} is the coefficient of static friction

or F = 51.4 N

Now, the box is sliding from the wall, so kinetic friciton comes into play. and the value of force is

F' = [tex] u_{K} [/tex] F
 or F' = 0.40 × 51.4 N

or F' = 20.56 N

The magnitude of the force (in N) pressing the block against the wall is  29.37 N.

What is frictional force?

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

Given is a 7.5-kg block is sliding down a wall with constant velocity. The coefficient of static friction between the block and the wall is μs = 0.70. The coefficient of kinetic friction between the block and the wall is μk = 0.40.

The friction force is given by

f = coefficient of friction x Normal force

The block is pressing the wall when gravitational acceleration is g = 9.79 m/s²

The frictional force is represented as

f =μk N =μk mg

where N is the normal force acting perpendicular to the wall due to contact between block and wall.

f =0.4 x 7.5 x  9.79

f = 29.37 N

Thus, the friction force is 29.37 N.

Learn more about friction force.

https://brainly.com/question/1714663


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