a 34.0 kg box initially at rest is pushed 4.40 m along arough, horizontal floor

Question

a 34.0 kg box initially at rest is pushed 4.40 m along arough, horizontal floor with a constant applied horizontal forceof 150 N. If the coefficient of friction between the box andthe floor is 0.300, find: a. the work done by the applied force b. the increase in internal energy in the box-floor system dueto friction c. the work done by the normal force d. the work done by the gravitational force e. the change in kinetic energy of the box f. the final speed of the box a 34.0 kg box initially at rest is pushed 4.40 m along arough, horizontal floor with a constant applied horizontal forceof 150 N. If the coefficient of friction between the box andthe floor is 0.300, find: a. the work done by the applied force b. the increase in internal energy in the box-floor system dueto friction c. the work done by the normal force d. the work done by the gravitational force e. the change in kinetic energy of the box f. the final speed of the box

Explanation / Answer

           Given that the mass of the box is m = 34.0 kg          The distance travelled is d = 4.40 m            The applied force is F = 150 N             Thecoefficient of friction between the box and the floor is = 0.300         --------------------------------------------------------------------------------------            The work done by the applied force is W1 = F*d                                                                              =150N * 4.40m                                                                             = ------------- J             The increase in the internal energy is due to friction (work doneby the friction )                                                                        W2 = f*d                                                                                =-*mg*d                                                                               = ------------------ J           Since there is no displancement along normal forcedirection then the work done is zero due to normal force.                 Since there isno displancement along gravitational forcedirection then the work done is zero due to normal force.                 From thework energy theorem the work done = change in kineticenergy                                                   (F- f)*d    = change in kinetic energy                                               ( F - mg)*d = change inkinetic energy                                                                       (F - f)*d =(1/2)mV2 -(1/2)mU2                                                                (F- mg)*d = (1/2)mV2- 0          (since initial velocity U = o)                                                                  V = [ 2*(F/m - g)*d ] 1/2                                                                      = ---------- m/s                                                                      = ---------- m/s                    Thisis the final speed of the box                                                                                                                                   

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