The answer isn't -9.9452 !
A 1210 kg roller coaster car is moving 7.33 m/s. As it approaches the station, brakes slow it down to 1.88 m/s over a distance of 5.29 m. How much work did the breaks do?
Fdcos(theta)
+37084 The work done on an object by a net force equals the change in kinetic energy of the object:
1/2mv^2 originally was (1/2) 1210 (7.33)^2 = 32505.98 j
later was (1/2) 1210 (1.88)^2 = 2138.31 j
Work = KEf - KEo = work done = -30367.67 j
+37084 The work done on an object by a net force equals the change in kinetic energy of the object:
1/2mv^2 originally was (1/2) 1210 (7.33)^2 = 32505.98 j
later was (1/2) 1210 (1.88)^2 = 2138.31 j
Work = KEf - KEo = work done = -30367.67 j
