Work, Energy, and Power Name

½•(4678 kg)•(17.1 m/s)² + F•(13.6 m)•cos(180°) = ½•(4678 kg)•(2.2 m/s)²

683946.9 … J - 11320.7 … J = F•(13.6 m)

F = 49500 N (49457.8 … N)
A catcher's mitt recoils a distance of 12.9 cm in bringing a 142-gram baseball to a stop. If the applied force is 588 N, then what was the speed of the baseball at the moment of contact with the catcher's mitt?
KE_i + W_ext = 0 J

½•(0.142 kg)•v_i² + (588 N)•(0.129 m)•cos(180°) = 0 J

(0.071 kg)•v_i² = 75.852 J

v_i = 32.7 m/s
An unknown force is applied to a 12 kg mass. The force acts at an angle of 30 degrees above the horizontal. Determine the force acting if the force acts for a horizontal displacement of 22 meters and increases the 12 kg mass's speed from 11 m/s to 26 m/s.
KE_i + W_ext = KE_f

½•(12 kg)•(11 m/s)² + F•(22 m)•cos(0°) = ½•(12 kg)•(26 m/s)²

F•(22 m) = 4056 J - 726 J

F= 150 N (151.36 … N)
A physics teacher exerts a force upon a 3.29-kg pile of snow to both lift it and set it into motion. The snow leaves the shovel with a speed of 2.94 m/s at a height of 0.562 m. Determine the work done upon the pile of snow.
W_ext = KE_f + PE_f

W = ½•(3.29 kg)•(2.94 m/s)² + (3.29 kg)•(9.8 N/kg)•(0.562 m)

W = 14.218 … J + 18.120 … J

W = 32.3 J
A 250.-gram cart starts from rest and rolls down an inclined plane from a height of 0.541 m. Determine its speed at a height of 0.127 m above the bottom of the incline.
PE_i = KE_f + PE_f

(0.250 kg)•(9.8 N/kg)•(0.541 m) = ½•(0.250 kg)•v_f² + (0.250 kg)•(9.8 N/kg)•(0.127 m)

1.325 … J - 0.311 … J = ½•(0.250 kg)•v_f²

v_f = 2.85 m/s
A 4357-kg roller coaster car starts from rest at the top of a 36.5-m high track. Determine the speed of the car at the top of a loop that is 10.8 m high.
KPE_i = KE_f + PE_f

(4357 kg)•(9.8 N/kg)•(36.5 m) = ½•(4357 kg)•v_f² + (4357 kg)•(9.8 N/kg)•(10.8 m)