In a horse cart system when would there be the smallest amount of acceleration
1 answer:
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Answer:
1) a. 52.41 m/s
b. The skier will be going 15.35 m/s slower
2) 103.68 m
3) 35,127 J
4) a. 88.825 kJ
(b) 16.36 %
5) 3,071.12 J
Explanation:
1) a. The given height of the hill, h = 140.0 m
The mass of the skier at the top of the hill, m = 85.0 kg
The acceleration due to gravity, g = 9.81 m/s²
The initial potential energy, P.E of the skier = m×g×h = 85.0×140.0×9.81 = 116739 J
From the principle of conservation of energy, we have;
The potential energy, P.E. lost = The gain in kinetic energy, K.E.
m×g×h = 1/2×m×v²
116739 J = 1/2×85.0×v²
v² = 116739/(1/2*85.0)= 2746.8 m²/s²
v = √(2746.8 m²/s²) = 52.41 m/s
b. From 70 m up, we have;
The initial potential energy, P.E., of the skier is now = 85.0×70×9.81 = 58,369.5 J
The potential energy, P.E. lost = The gain in kinetic energy, K.E.
58,369.5 J = 1/2×85.0×v²
v² = 58,369.5/(1/2*85.0) = 1373.4 m²/s²
v = 37.06 m/s
The skier will be going 52.41 - 37.06 = 15.35 m/s slower
The skier will be going 15.35 m/s slower
2) From the principle of conservation of energy, the amount of work done (energy used) = The (potential) energy gained by the load
The amount of work done by the electric hoist = 356,000 J
The mass of the load = 350.0 kg
The height to which the load is raised = h
The potential energy gained by the load = m×g×h = 350.0×9.81×h
356,000 J = 350.0×9.81×h
h = 356,000/(350.0*9.81) = 103.68 m
The height to which the load is lifted= 103.68 m
3) The initial potential energy of the roller coaster cart = 600*35.0*9.81 = 206010 J
The final potential energy = 600*28.0*9.81= 164808 J
The velocity at point 3 = 4.5 m/s
The kinetic energy at point 3 = 1/2*600*4.5^2 = 6075 J
The total energy at point 3 = 164808 + 6075 = 170,883 J
The energy loss = The initial potential energy at point 1 - Total energy at point 3
The energy loss = 206010 - 170,883 = 35,127 J
The heat energy due to friction that must have been produced between points 1 and 3 = 35,127 J
4) a. The heat energy absorbed = mass × specific heat capacity for water, × Temperature change
The mass of the water = 2.5×10² g = 0.25 kg
= 4,180 J/(kg·°C)
Initial temperature = 10.0°C
Final temperature = 95°C
The temperature change = 95.0°C - 10.0°C = 85.0°C
The heat energy absorbed = 0.25*4,180* 85 = 88,825 J = 88.825 kJ
(b) The percentage efficiency = (Heat absorbed/(Heat supplied)) × 100
The heat supplied = 543 kJ
The efficiency = (88.825/543) × 100 = 16.36 %
5) The mass of the box = 115 kg
Force acting on the rope = 255 N
The angle of inclination of the force to the horizontal = 24.5°
The distance the box is displaced = 15.0 m to the right
The work done = Force applied × distance moved in the direction of the force
The work done = Force applied × distance moved in the direction of the force
Given that the load moves a distance 15.0 m to the right,we have;
The component of the force acting in the direction of the movement of the load (to the right) is 225 × cos(24.5°) = 204.74 N
The work done = 204.7*15 = 3071.12 J
The amount of work done = 3,071.12 J