Question

A 20 kg box s initially at rest. Two forces drag. it across the floor, F_1 = 35 N i and F_2 = 35 N j, which cause a displacement of 3.5 m Imagine the x-y plane lies on the floor. (a) Draw the displacement vector and show (no need to calculate) the direction in which the package moves? (b) How much work is done by each force individually (F_1 and F_2)? (c) The coefficient of kinetic friction between the box and the floor is mu = 0.25. Using the work on the box, find the speed of the box after it has been dragged 3.5 m. (do not use kinematics to solve this problem, use only work and energy principles)

Answer 1

Mass of box = 20 kg Resultant of F_1 and F_2 is 35 squareroot 2 N making an angle of tan theta = F_2/F_1 = 1 or theta = 45 degree with x - direction Thus block will move in direction of resultant force so package move along Vector r. work done by F_1 = (3 5)(3 5) = 122.5 J work done by F_2 = (35)(3.5) = 122.5 J \r\n Work done by function on box W_f = -fd = -mu_k mg d mu_k: coefficient of friction d: distance moved Resultant of force F_1 and F_2 is squareroot 35^2 + 35^2 = 35 squareroot 2 N work done by resultant of F_1, and F_2 over a distance of 3.5m W = (35 squareroot 2)(3.5) Thus total work done on box is equal to change in its kinetic energy W + w_f = 1/2 mv^2 (35 squareroot 2)(3.5) - (0.25)(20)(9.81)(3.5) = (0.5)(20)v^2 v = 0.395 m/s