Question

Find the force acting on it and the acceleration. 15 points. Will give brainliest!

Answer 1

Answer:

$F_{net} = 1905 N$

$\boxed{a = 7.62 m/s^2}$

Explanation:

For net force:

Firstly, finding the force of friction:

$F_{k} =$ $(micro)_{k} m g cos \theta$

Where $(micro)_{k}$ is the coefficient of friction, m =  250 kg and g = 9.8 m/s²

$F_{k} = (0.17)(25)(9.8) cos (52)\\F_{k} = (41.65)(0.616)\\F_{k} = 25.59 \ N$

Now, Finding of the pulling box:

$F = mgsin\theta$

$F = (250)(9.8) Sin (52)\\F = 2450 * 0.788\\F = 1930.6$

So, The net Force is

$F _{net} = F - F_{k}$

$F_{net} = 1930.6 -25.6$

$F_{net} = 1905 N$

For Acceleration:

$F_{net} = m a$

For a, it is:

=> $a = \frac{F_{net}}{m}$

=> a = 1905 / 250

=> a = 7.62 m/s²