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anygoal [31]
3 years ago
9

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, cau

sing it to start sliding downward, starting from rest. If it starts 4.25 m 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 22.0 N?
Physics
1 answer:
eduard3 years ago
6 0

Answer:

The speed is   v =8.17 m/s

Explanation:

From the question we are told that

      The angle of slant is  \theta = 37.0^o

       The weight of the toolbox is  W_t = 92.0N

       The mass of the toolbox is m = \frac{92}{9.8} = 9.286kg

       The start point is  d = 4.25m from lower edge of roof

        The kinetic frictional force is  F_f = 22.0N

Generally the net work done on this tool box can be mathematically represented as

      Net \ work done  =  Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction

The workdone due to weigh is  =    mgsin \theta * d

 The workdone due to friction is  = F_f \ cos\theta  *   d

Substituting this into the equation for net workdone  

                 W_{net} = mgsin\theta  * d + F_f  \ cos \theta *d

      Substituting values

                  W_{net}  =  92 * sin (37)  * 4.25 + 22 cos (37) * 4.25

                          = 309.98 J

 According to work energy theorem

             W_{net} = \Delta Kinetic \ Energy

              W_{net} = \frac{1}{2} m (v - u)^2

From the question we are told that it started from rest so  u = 0 m/s

              W_{net} = \frac{1}{2} * m v^2

Making v the subject

               v = \sqrt{\frac{2 W_{net}}{m} }

Substituting value

              v = \sqrt{\frac{2 * 309.98}{9.286} }

             v =8.17 m/s

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