5. QC Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0° incline. The coefficient of kinetic friction between the block and the incline is ur = 0.436. Determine (a) the work done by the force of gravity, (b) the work done by the friction force between block and incline, and (c) the work done by the normal force. (d) Qualitatively, how would the answers change if a shorter ramp at a steeper angle were used to span the same vertical height? Explanation of Solution Given Info: The mass of the block is 5.00 kg. The length of the surface that the block slides is 2.50 m. The inclination of the surface is 30.0°. L= 2.50 m in »,-»| mgv 30.0° Formula to calculate the work done is, W = (F cos 0) d (I) O is the angle between the force vector and the displacement vector • Fis the force • dis the magnitude of displacement Since, the gravitational force acting on the object is, F = mg • m is the mass of the block • gis acceleration due to gravity Thus, equation (I) gives, W = (mg cos 0) d Substitute 5.00 kg for m, 9.8 m/s² for g, 30.0° for 0 and 2.50 m for d to find the work done, W = (5.00 kg) (9.8 m/s²) (sin 30.0°) (2.50 m) = 64.3 J Thus, the work done is 64.3 J. Conclusion: The total work done by the factor of gravity is 64.3 J.

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