A boxed 10.0-kg computer monitor is dragged by friction 5.50 m upward along a conveyor belt inclined at an angle of 36.9° above the horizontal. If the monitor’s speed is a constant 2.10 cm/s, how much work is done on the monitor by (a) friction, (b) gravity, and (c) the normal force of the conveyor belt?

Solution 11E Step 1: The force acting on the box will be, F = - mg sin Step 2: Mass of the box, m = 10 kg 2 Acceleration due to gravity, g = 9.8 m/s Angle with the horizontal, = 36.9° 2 Therefore, force to lift the box through the inclined plane, F = - 10 kg × 9.8 m/s × sin 36.9° sin 36.9 = 0.60 2 F = - 10 kg × 9.8 m/s × 0.60 = - 58.8 N