A 5.00-kg box sits at rest at the bottom of a ramp that is 8.00 m long and that is inclined at 30.0° above the horizontal. The coefficient of kinetic friction is ???k = 0.40 and the coefficient or static friction is ???s = 0.50. What constant force ?F?, applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of 4.00 s?
Solution 64P Step 1: Figure 1 represents the condition of the box moving in an inclined plane. Provided, = 30°, m = 5 kg, t = 4 s, and = 0.40. We know that, g = 9.8 m/s 2 k Step 2: As per the diagram, we can see that, the component, mg cos and normal component, - mg cos are cancelled each other. The component, F = mg sin is due to the gravitational force as shown in the diagram. g 2 Therefore, F = 5 g × 9.8 m/s × sin 30 = 49 N × ½ Fg 24.5 N The force due to kinetic friction, f = mg cos at an inclined plane. k 2 3 f = 5 kg × 9.8 m/s × cos 30 = 0.4×49 N × 2 f = 0.4×24.5 3 N = 0.4×42.44 N = 16.976 N We can take the resultant since, both of these forces are acting towards the same direction. Finclined24.5 N + 16.976 N = 41. 476 N