The figure is not visible. Consider the problem data: `a, mu_k`

The coefficient of kinetic friction is `mu_k = (f_k)/n`

Substitute 0.3 for `mu_k => 0.3 = (f_k)/n => f_k = 0.3*n`

Applying Newton's second law:

`Sigma (F_x): F*cos theta - f_k = m*a`

Substitute 2.9 for `a` =>` F*cos theta - 0.3*n = 2.9*m`

`Sigma` `(F_y): F*sin theta + n - m*g= 0 => n = m*g - F*sin theta`

Substitute `m*g - F*sin theta` for `n` in equation `F*cos theta - 0.3*n = 2.9*m` :

`F*cos theta - 0.3*m*g - F*sin theta = 2.9*m`

`F*cos theta - F*sin theta = m*(0.3*g + 2.9)`

Put `g = 10m/s^2`

`F*(cos theta - sin theta) = m*(3 + 2.9) => (cos theta - sin theta) =5m/F`

Use `cos theta = sqrt(1 - sin^2 theta)`

`sqrt(1 - sin^2 theta) - sin theta = 5m/F`

`(sqrt(1 - sin^2 theta) - sin theta)^2 = 25(m/F)^2`

Square the left binomial:

`1 - sin^2 theta - 2sintheta*cos theta + sin^2 theta = 25(m/F)^2`

Use `sin 2theta` instead of `2sintheta*cos theta`

`1 - sin 2theta = 25(m/F)^2 => 1 - 25(m/F)^2 = sin 2 theta => 2theta = arcsine (1 - 25(m/F)^2) => theta = (arcsine (1 - 25(m/F)^2) )/2`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.