a) -442.6 J
b) 8.8 m/s
Explanation:
a)
The work done by the friction force is equal to:
[tex]W_f=-F_f d[/tex]
where
[tex]F_f = \mu_k mg cos \theta[/tex] is the frictional force on the box, where
[tex]\mu_k = 0.48[/tex] is the coefficient of friction
m = 21 kg is the mass of the box
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]\theta[/tex] is the angle of the ramp
The negative sign is due to the fact that the frictional force is opposite to the motion of the box
d is the length of the ramp
So the work can be rewritten as
[tex]W_f=-\mu_k mg cos \theta d[/tex]
Here we know that the grade of the ramp, which is the ratio between height and horizontal length, is 0.7:
[tex]grade=\frac{h}{L}=0.7[/tex]
This means that
[tex]tan \theta=0.7[/tex]
So we find the angle:
[tex]\theta=tan^{-1}(0.7)=35.0^{\circ}[/tex]
We also know that the height of the ramp is
h = 3.14 m
So we can find the length of the ramp:
[tex]d=\frac{h}{sin \theta}=\frac{3.14}{sin 35^{\circ}}=5.47 m[/tex]
Therefore, the work done by friction is
[tex]W_f=-(0.48)(21)(9.8)(cos 35^{\circ})(5.47)=-442.6 J[/tex]
b)
The final speed of the box can be found by using the law of conservation of energy. In fact, the initial kinetic energy of the box (at the bottom) + the work done by friction must be equal to the final kinetic energy of the box, as it reaches the ground after leaving the platform.
So we can write:
[tex]KE_i+W_f=KE_f\\\frac{1}{2}mu^2 + W_f = \frac{1}{2}mv^2[/tex]
where:
u = 10.9 m/s is the initial speed of the box
m = 21 kg is the mass of the box
[tex]W_f=-442.6 J[/tex] is the work done by friction
[tex]v[/tex] is the final speed of the box as it reaches the ground
Solving for v, we find:
[tex]v=\sqrt{u^2+\frac{2W_f}{m}}=\sqrt{10.9^2+\frac{2(-442.6)}{21}}=8.8 m/s[/tex]
