Given :
Mass of box , m = 250 kg.
Force applied , F = 285 N.
The value of the incline angle is 30°.
the coefficient of dynamic friction is [tex]\mu=0.72[/tex] .
To Find :
The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.
Solution :
Net force applied in box is :
[tex]F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N[/tex]
Acceleration , [tex]a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2[/tex].
By equation of motion :
[tex]v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s[/tex]
Therefore, the speed of box is 12.04 m/s.
Hence, this is the required solution.
