While a roofer is working on a roof that slants at 43.0 above the horizontal, he accidentally nudges his 93.0N toolbox, causing it to start

Question

While a roofer is working on a roof that slants at 43.0 above the horizontal, he accidentally nudges his 93.0N toolbox, causing it to start sliding downward, starting from rest.Part AIf it starts 4.60m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 21.0N ?

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Philomena 3 years 2021-08-26T19:23:22+00:00 1 Answers 18 views 0

Answers ( )

  1. Nick
    0
    2021-08-26T19:25:16+00:00
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    Answer:

    6.41m/s

    Explanation:

    #There’s kinetic friction force acting on the block W_{other}\neq 0

    So the work energy theorem is expressed as:

    K_1+U_{grav,1}+W_{other}=k_2+U_{grav,2}\\

    #First we calculate the work done by friction force by substituting for our f_k, s=d values into W=F_s:

    W=F_s\\\\=-21N\times4.6m\\\\=-96.60J#-ve, friction is opposite to displacement.

    We then calculate the height from point1 to point2 as:

    h=dsin\theta\\=4.6sin 43\textdegree\\\\=3.14m

    Mass of toolbox is calculated as follows:

    m=F_g/g\\=93.0/9.8\\\\=9.49kg

    #Substitute given values from the question in KE=0.5mv^2  to get the kinetic energy of the toolbox:

    K_1=0.5\times 9.49\times0\\\\=0

    #Substitute given values from the question into U_{grav}=F_gy to get the gravitational potential energy:

    U_{grav}=F_gy\\\\U_{grav,1}=93\times 3.14\\\\=292.02J

    At point 2, the kinetic energy is:

    K_2=0.5(9.49)v_2^2=4.75v_2^2\\U_{grav,2}=0\\

    We then substitute the calculate values for energy quantities into K_1+U_{grav,1}+W_{other}=k_2+U_{grav,2}\\

    0+292.02J-96.60j=4.75v_2^2+0\\\\v_2^2=41.14\\\\v_2=6.41m/s

    Hence, toolbox will be moving at a velocity of 6.41m/s as it reaches the edge of the roof.

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