Answer:
21.2 degrees
Explanation:
Let gravitational acceleration constant g = 9.81 m/s2 directed downward. We can calculate the g component that is parallel to the 25 degree incline:
[tex]gsin25^0 = 9.81sin25^0 = 4.15 m/s^2[/tex]
This parallel component would produce an acceleration of the block. But since the net acceleration is only 0.7 m/s2, there's a friction acceleration that hinders g parallel. This acceleration can be calculated by
[tex]4.15 - a_f = a = 0.7[/tex]
[tex]a_f = 4.15 - 0.7 = 3.45 m/s^2[/tex]
The force of friction would be
[tex]F_f = a_fm = 250 * 3.45 = 816.5 N[/tex]
Friction is a product of normal force and its coefficient
[tex]F_f = \mu N = \mu mgcos25^0 = \mu 250*9.81*cos25^0 = 2222.72 \mu[/tex]
[tex]\mu = F / 2222.72 = 816.5 / 2222.72 = 0.388[/tex]
For the crate to slide down at constant speed, the net acceleration must be 0. In other words, the parallel g must be the same as acceleration caused by friction. Let this incline angle be α
[tex]g sin\alpha = a_f = F_f/m = \frac{\mu N}{m} = \frac{\mu mgcos\alpha}{m} = \mu g cos\alpha[/tex]
[tex]sin\alpha = \mu cos\alpha[/tex]
[tex]\frac{sin \alpha}{cos \alpha} = \mu[/tex]
[tex]tan\alpha = \mu = 0.388[/tex]
[tex]\alpha = tan^{-1} 0.388 = 0.37 rad = 0.37*180/\pi \approx 21.2^0[/tex]
