One person can hold mutiple types of authority at a time T/F
1 answer:
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Answer:
a) 855.41J
b) 855.41J
Explanation:
The force equation is
N - mg - F sinθ = 0
N = mg + F sinθ
The frictonal force is =
substitute the value of the normal force in the above equation
= 151.96N
a) work done = F × d
= (151.96) × (6.5) cos30°
= 855.41 J
b) The increase in thermal energy of the block floor system will be equal to the work done by the worker
work done = F × d
= (151.96) × (6.5) cos30°
= 855.41 J
Answer:
1. The work done on the cube during the time the cube is in contact with the spring is 0.8023705 J
2. The speed of the cube at the instant just before the sliding cube leaves the ramp is approximately 31.5 cm/s
Explanation:
The given parameters of the Rube Goldberg machine are;
The distance from the free end of the spring to the top of the ramp, d = 17.0 cm = 0.17 m
The mass of the small cube to be launched, m = 119.0 g = 0.119 kg
The spring constant of the spring, k = 461.0 N/m
The angle of elevation of the ramp to the horizontal, θ = 50.0°
The coefficient of static friction of the wood, = 0.590
The coefficient of dynamic friction of the wood, = 0.470
The velocity of the cube at the top of the ramp, v = 45.0 cm/s = 0.45 m/s
The amount by which the cube is compressed, x = 5.90 cm = 0.059 m
The work done on the cube during the time the cube is in contact with the spring = The energy of the spring, E = (1/2)·k·x²
∴ E = (1/2) × 461.0 N/m × (0.059 m)² = 0.8023705 J
The work done on the cube during the time the cube is in contact with the spring= E = 0.8023705 J
2. The frictional force, =
·m·g·cos(θ)
∴ = 0.470 × 0.119 × 9.8 × cos(50) ≈ 0.35232 N
The work loss to friction, W = × d
∴ W = 0.35232 N × 0.17 m ≈ 0.05989 J
The work lost to friction, W ≈ 0.05989 J
The potential energy of the cube at the top of the ramp, P.E. = m·g·h
∴ P.E. = 0.119 kg × 9.8 m/s² × 0.17 m × sin(50°) ≈ 0.151871375 J
By conservation of energy principle, the Kinetic Energy of the cube at the top of the ramp, K.E. = E - W - P.E.
∴ K.E. = 0.8023705 J - 0.05989 J - 0.151871375 J ≈ 0.590609125 J
K.E. = (1/2)·m·v²
Where;
v = The speed of the cube at the instant just before the sliding cube leaves the ramp
∴ K.E. = (1/2) × 0.119 kg × v² ≈ 0.590609125 J
v² ≈ 0.590609125 J/((1/2) × 0.119 kg) ≈ 9.92620378 m²/s²
v = √(9.92620378 m²/s²) ≈ 3.15058785 m/s ≈ 31.5 cm/s
The speed of the cube at the instant just before the sliding cube leaves the ramp, v ≈ 31.5 cm/s.