A 6 N horizontal force F pushes a block weighing 10.0 N against a vertical wall

Question

A 6 N horizontal force F pushes a block weighing 10.0 N against a vertical wall (Fig. 6-21). The coefficient of static friction between the wall and the block is 0.57, and the coefficient of kinetic friction is 0.37. Assume that the block is not moving initially. (let up and to the right be positive dirrections) Figure 6-21 (a) Will the block move? no yes (b)What is the frictional force on the block? N (c)What is the acceleration of the block?(the answer is a negative number) m/s2 (d)What would the Frictional force be if the horizontal force was 540N?

Explanation / Answer

Friction = x N = (0.57) x (6) = 3.42 Newtons (acting up)

Gravity = m x g = 10.0 Newtons (acting down)

(horizontal forces cancel each other out)

Therefore, the block moves down, as even the maximum static friction can't keep it up

Net force = 10 - (0.37)(6) = 7.78 Newtons

Net acceleration = (7.78/(10/9.8)) = 7.6244 m/sec2

a. Yes

b. (0.37) x (6) = 2.22 Newtons

c. 7.64244 m/sec2

d. It would be   x N = (540) x (6) = 3240 Newtons

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