From Newton's second law of motion, the acceleration of the crate is 0.84m/[tex]s^{2}[/tex]
Given that a 25 kg crate is pushed horizontally with a force of 70N and the coefficient of friction is 0.20
To calculate the acceleration of the crate, We will consider the net force acting on the crate. The net force will be sum of the applied force F and the frictional force [tex]F_{r}[/tex].
Let us first calculate the frictional force [tex]F_{r}[/tex]. That is,
[tex]F_{r}[/tex] =- μN
[tex]F_{r}[/tex] = μmg
[tex]F_{r}[/tex] = 0.2 x 25 x 9.8
[tex]F_{r}[/tex] = 49N
Using Newton's second law,
F = ma
F - [tex]F_{r}[/tex] = ma
70 - 49 = 25 a
21 = 25a
a = 21 / 25
a = 0.84 m/[tex]s^{2}[/tex]
Therefore, the acceleration of the crate is 0.84m/[tex]s^{2}[/tex]
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