Problem 1

Distance it falls is D = (1/2)gt² = (1/2)(9.8)(0.45)² = 0.992 m.
Initially the mass is at rest, velocity = 0, kinetic energy = 0.
Potential energy = mgD = (0.5)(9.8)(0.992) = 4.86 J.
Velocity after time t is given by v = gt. v = (9.8)(0.45) = 4.41 m/s.
Kinetic energy at the ground = (1/2)mv² = (1/2)(0.5)(4.41)² = 4.86 J.
Potential energy at the ground = mgD = 0.
Total energy = PE + KE = 4.86 J.

Problem 2

Distance it falls is D = (1/2)gt² = (1/2)(9.8)(0.557)² = 1.52 m.
Initially the mass is at rest, velocity = 0, kinetic energy = 0.
Potential energy = mgD = (0.2)(9.8)(1.52) = 2.98 J.
Velocity after time t is given by v = gt. v = (9.8)(0.557) = 5.46 m/s.
Kinetic energy at the ground = (1/2)mv² = (1/2)(0.2)(5.46)² = 2.98 J.
Potential energy at the ground = mgD = 0.
Total energy = PE + KE = 2.98 J.

Problem 3

Distance it falls is D = (1/2)gt² = (1/2)(9.8)(0.2)² = 0.196 m.
Initially the mass is at rest, velocity = 0, kinetic energy = 0.
Potential energy = mgD = (1.2)(9.8)(0.196) = 2.305 J.
Velocity after time t is given by v = gt. v = (9.8)(0.2) = 1.96 m/s.
Kinetic energy at the ground = (1/2)mv² = (1/2)(1.2)(1.96)² = 2.305 J.
Potential energy at the ground = mgD = 0.
Total energy = PE + KE = 2.305 J.

Problem 4

Distance it falls is D = (1/2)gt² = (1/2)(9.8)(0.3)² = 0.441 m.
Initially the mass is at rest, velocity = 0, kinetic energy = 0.
Potential energy = mgD = (0.4)(9.8)(0.441) = 1.729 J.
Velocity after time t is given by v = gt. v = (9.8)(0.3) = 2.94 m/s.
Kinetic energy at the ground = (1/2)mv² = (1/2)(0.4)(2.94)² = 1.729 J.
Potential energy at the ground = mgD = 0.
Total energy = PE + KE = 1.729 J.

Problem 5

Since the masses of hot and cold water are equal, the final temperature is just half way between 80 and 40 degrees. T_final = 60 degrees.
Energy lost by hot water = cm(TempChange) = (1)(100)(80 - 60) = 2000 cal. [Specific heat, c = 1 for water.]
Energy gained by cold water = cm(TempChange) = (1)(100)(60 - 40) = 2000 cal.

Problem 6

Energy lost by hot water = cm(TempChange) = (1)(100)(70 - 50) = 2000.
Energy gained by cold water = cm(TempChange) = (1)(200)(50 - 40) = 2000.

Problem 7

Energy lost by hot water = cm(TempChange) = (1)(300)(60 - 49) = 3300 cal.
Energy gained by cold water = cm(TempChange) = (1)(100)(49 - 20) = 2900 cal.
Heat lost = 3300 - 2900 = 400 cal.

Problem 8

Energy lost by aluminum = cm(TempChange) = (0.217)(30)(80 - 21.89) = 378 cal.
Energy gained by water = cm(TempChange) = (1)(200)(21.89 - 20) = 378 cal.

Problem 9

Energy lost by copper = cm(TempChange) = (0.093)(30)(80 - 22.9) = 159.3 cal.
Energy gained by water = cm(TempChange) = (1)(150)(22.9 - 22) = 135 cal.
Heat lost = 159.3 - 135 = 24.3 cal.