Given:
The mass of the granite block is: m = 1400 kg
The granite block moves with the constant speed of: v = 1.34 m/s
The distance d1 is: d1 = 40 m
The distance d2 is: d2 = 30 m
The coefficient of kinetic friction of the block and the incline is: μ = 0.40
To find:
The power of the cable's force applied to the block.
Explanation:
The forces acting on the block can be resolved into its components as:
The tension in the rope is given as:
[tex]T=mgsinθ+f[/tex]
Here, f is the force of friction, which is given as:
[tex]f=\mu N[/tex]
The normal force is balanced by the mgcosθ, thus,
[tex]N=mgcosθ[/tex]
Thus, the tension in the string is given as:
[tex]T=mgsinθ+\mu mgcosθ[/tex]
The angle θ made by inclined with the horizontal can be calculated as:
[tex]\begin{gathered} θ=tan^{-1}(\frac{30\text{ m}}{40\text{ m}}) \\ \\ θ=tan^{-1}(0.75) \\ \\ θ=36.87\degree \end{gathered}[/tex]
The tension in the string can be calculated as:
[tex]\begin{gathered} T=1400\text{ kg}\times9.8\text{ m/s}^2\times sin(36.87\degree)+0.40\times1400\text{ kg}\times9.8\text{ m/s}^2\times cos(36.87\degree) \\ \\ T=12622.41\text{ kg.m/s}^2 \end{gathered}[/tex]
Now the power P of the cable's force applied to the block can be calculated as:
[tex]P=Tv[/tex]
Substituting the values in the above equation, we get:
[tex]\begin{gathered} P=12622.41\text{ kg.m/s}^2\times1.34\text{ m/s} \\ \\ P=16914.03\text{ kg.m}^2\text{/s}^3 \\ \\ P=16914.03\text{ J/s} \\ \\ P=16914.03\text{ W} \end{gathered}[/tex]
Final answer:
The power of the cable's force applied to the box is 16914.03 W.
