A block whose weight is 45 N rests on a horizontal table. A horizontal force of 36 N is applied to the block. The coefficient of kinetic and static 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 blocks acceleration?

This is less than the 36 N applied, so the block will move. Since the block is moving, kinetic friction takes over. To find the block's acceleration, use Newton's second law:

∑F = ma

F − N μk = ma

36 N − (45 N) (0.65) = (45 N / 9.8 m/s²) a

6.75 N = 4.59 kg a

a = 1.47 m/s²

Rounded to two significant figures, the block's acceleration is 1.5 m/s².

Usually the coefficient of static friction is greater than the coefficient of kinetic friction. You might want to double check the problem statement, just to be sure.

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-11-03T15:08:46+01:00

The block will move since the horizontal force is greater than the static frictional force.

The acceleration of the block under the influence of the force is 1.47 m/s².

The given parameters;

weight of the block, W = 45 N
horizontal force, F = 36 N
coefficient of kinetic friction = 0.65
coefficient of static friction, = 0.42

The mass of the block is calculated as follows;

[tex]W = mg\\\\m = \frac{W}{g} \\\\m = \frac{45}{9.8}\\\\m = 4.59 \ kg[/tex]

The static frictional force on the block is calculated as follows;

[tex]F_s = \mu_s F_n\\\\F_s = \mu_ s W\\\\F_s = 0.42 \times 45\\\\F_s = 18.9 \ N[/tex]

The block will move since the horizontal force is greater than the static frictional force.

The acceleration of the block is calculated as follows;

[tex]F- \mu_ k F_n = ma\\\\36 - (0.65 \times 45) = 4.59a\\\\ 6.75 = 4.59a\\\\a = \frac{6.75}{4.59} \\\\a = 1.47 \ m/s^2[/tex]

Thus, the acceleration of the block under the influence of the force is 1.47 m/s².

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