A 150 N Sled is pulled up a 28 Degree slope at a constantSpeed by a 100 N Force

Question

A 150 N Sled is pulled up a 28 Degree slope at a constantSpeed by a 100 N Force that is Parallel to the Slope.

1. What is the coefficient of Kinetic Friction between the Sled andthe Slope?

2. The Sled is the Released when near the top of the hill. What isthe Acceleration of the sled as it slides down the hill?

Please Show All WORK and ALL FORMULAS used to get the answer Which Answer is correct? We can write when sled pulling upward direction
         100N - mgsin(28) -mgcos(28)k = 0  
          k = [100N-mgsin(28)]/mgcos(28)
                =[100N-150N*sin(28)]/(150N*cos(28))                 =0.223    

The acceleration be a, we have:          mgsin(28)-mgcos(28) = ma           a =gsin(28) - gcos(28)             = 2.67m/s2 ________________________________________________________________________________________ OR weight of sled W = 150 N angle = 28 degrees force F = 100 N (1) . net force parallel to slope F ' = 100 -W cos 28 Since it moves with constant speed , F =0                                     W cos 28 = 100                                                   = 0.755 the coefficient of Kinetic Friction between the Sled andthe Slope = 0.755 (2). Initial velocity u = 0 Accleration   a = g sin 28 - g cos28                       = 4.6 -0.8829                       = 3.717 m / s ^ 2 A 150 N Sled is pulled up a 28 Degree slope at a constantSpeed by a 100 N Force that is Parallel to the Slope.

1. What is the coefficient of Kinetic Friction between the Sled andthe Slope?

2. The Sled is the Released when near the top of the hill. What isthe Acceleration of the sled as it slides down the hill?

Please Show All WORK and ALL FORMULAS used to get the answer Which Answer is correct? We can write when sled pulling upward direction
         100N - mgsin(28) -mgcos(28)k = 0  
          k = [100N-mgsin(28)]/mgcos(28)
                =[100N-150N*sin(28)]/(150N*cos(28))                 =0.223    

The acceleration be a, we have:          mgsin(28)-mgcos(28) = ma           a =gsin(28) - gcos(28)             = 2.67m/s2 ________________________________________________________________________________________ OR weight of sled W = 150 N angle = 28 degrees force F = 100 N (1) . net force parallel to slope F ' = 100 -W cos 28 Since it moves with constant speed , F =0                                     W cos 28 = 100                                                   = 0.755 the coefficient of Kinetic Friction between the Sled andthe Slope = 0.755 (2). Initial velocity u = 0 Accleration   a = g sin 28 - g cos28                       = 4.6 -0.8829                       = 3.717 m / s ^ 2 We can write when sled pulling upward direction
         100N - mgsin(28) -mgcos(28)k = 0  
          k = [100N-mgsin(28)]/mgcos(28)
                =[100N-150N*sin(28)]/(150N*cos(28))                 =0.223    

The acceleration be a, we have:          mgsin(28)-mgcos(28) = ma           a =gsin(28) - gcos(28)             = 2.67m/s2 ________________________________________________________________________________________ OR weight of sled W = 150 N angle = 28 degrees force F = 100 N (1) . net force parallel to slope F ' = 100 -W cos 28 Since it moves with constant speed , F =0                                     W cos 28 = 100                                                   = 0.755 the coefficient of Kinetic Friction between the Sled andthe Slope = 0.755 (2). Initial velocity u = 0 Accleration   a = g sin 28 - g cos28                       = 4.6 -0.8829                       = 3.717 m / s ^ 2 The acceleration be a, we have:          mgsin(28)-mgcos(28) = ma           a =gsin(28) - gcos(28)             = 2.67m/s2 ________________________________________________________________________________________ OR weight of sled W = 150 N angle = 28 degrees force F = 100 N (1) . net force parallel to slope F ' = 100 -W cos 28 Since it moves with constant speed , F =0                                     W cos 28 = 100                                                   = 0.755 the coefficient of Kinetic Friction between the Sled andthe Slope = 0.755 (2). Initial velocity u = 0 Accleration   a = g sin 28 - g cos28                       = 4.6 -0.8829                       = 3.717 m / s ^ 2 weight of sled W = 150 N angle = 28 degrees force F = 100 N (1) . net force parallel to slope F ' = 100 -W cos 28 Since it moves with constant speed , F =0                                     W cos 28 = 100                                                   = 0.755 the coefficient of Kinetic Friction between the Sled andthe Slope = 0.755 (2). Initial velocity u = 0 Accleration   a = g sin 28 - g cos28                       = 4.6 -0.8829                       = 3.717 m / s ^ 2

Explanation / Answer

                =0.223    

The acceleration be a, we have:          mgsin(28)-mgcos(28) = ma           a =gsin(28) - gcos(28)             = 2.67m/s2 The acceleration be a, we have:          mgsin(28)-mgcos(28) = ma           a =gsin(28) - gcos(28)             = 2.67m/s2

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