Answer:
6.41m/s
Explanation:
#There's kinetic friction force acting on the block [tex]W_{other}\neq 0[/tex]
So the work energy theorem is expressed as:
[tex]K_1+U_{grav,1}+W_{other}=k_2+U_{grav,2}\\[/tex]
#First we calculate the work done by friction force by substituting for our [tex]f_k, s=d[/tex] values into [tex]W=F_s[/tex]:
[tex]W=F_s\\\\=-21N\times4.6m\\\\=-96.60J[/tex]#-ve, friction is opposite to displacement.
We then calculate the height from point1 to point2 as:
[tex]h=dsin\theta\\=4.6sin 43\textdegree\\\\=3.14m[/tex]
Mass of toolbox is calculated as follows:
[tex]m=F_g/g\\=93.0/9.8\\\\=9.49kg[/tex]
#Substitute given values from the question in [tex]KE=0.5mv^2[/tex] to get the kinetic energy of the toolbox:
[tex]K_1=0.5\times 9.49\times0\\\\=0[/tex]
#Substitute given values from the question into [tex]U_{grav}=F_gy[/tex] to get the gravitational potential energy:
[tex]U_{grav}=F_gy\\\\U_{grav,1}=93\times 3.14\\\\=292.02J[/tex]
At point 2, the kinetic energy is:
[tex]K_2=0.5(9.49)v_2^2=4.75v_2^2\\U_{grav,2}=0\\[/tex]
We then substitute the calculate values for energy quantities into [tex]K_1+U_{grav,1}+W_{other}=k_2+U_{grav,2}\\[/tex]
[tex]0+292.02J-96.60j=4.75v_2^2+0\\\\v_2^2=41.14\\\\v_2=6.41m/s[/tex]
Hence, toolbox will be moving at a velocity of 6.41m/s as it reaches the edge of the roof.
