# Physics I: a 5.2kg box is sliding down a ramp to a loading dock. The coefficient of kinetic energy is given as 0.25. Find the normal force, friction force, and acceleration of the block from the...

Physics I: a 5.2kg box is sliding down a ramp to a loading dock. The coefficient of kinetic energy is given as 0.25. Find the normal force, friction force, and acceleration of the block from the following image.

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**Images:**

You need to remember that the coefficient of kinetic friction is defined as a ratio of force of friction to the normal force, such that:

`mu_k = F_f/F_n`

`F_f` represents the force of friction

`F_n` represents the normal force

Since the problem provides the coefficient of kinetic friction yields:

`0.25 = F_f/F_n`

Since the weight force is the only force that acts on the sliding box, you may write the equilibrium equation on y axis to evaluate the normal force, such that:

`sum F_y = 0: F_n - m*g*cos 30^o = 0 => F_n = m*g*cos 30^o`

`F_n = 5.2*9.8*sqrt3/2 => F_n = 44.1326 N`

Since `F_f = 0.25F_n` yields:

`F_f = 0.25*44.1326 => F_f = 11.03315N`

You need to write the equilibrium equation on x axis to evaluate the acceleration, such that:

`sum F_x = 0 => m*g*sin 30^o + F_f = m*a`

Factoring out `m` yields:

`m(g*sin 30^o + 0.25g*cos 30^o) = m*a`

Reducing duplicate factors yields:

`a = g(sin 30^o + 0.25*cos 30^o)`

`a = 9.8(1/2 + sqrt3/8)`

`a = 7.021 m*s^(-2)`

**Hence, evaluating the normal force, the friction force and acceleration, under the given conditions, yields `F_n = 44.1326 N, F_f = 11.03315N` andÂ **`a = 7.021 m*s^(-2).`