Answer:
V is approximately = 23m/s
Explanation:
Kinetic energy = ½ mv²
Where m= mass = 0.450kg
V= velocity =?
K. E = 119J
Therefore
K. E = ½ mv²
Input values given
119= ½ × 0.450 × v²
Multiply both sides by 2
119 ×2 = 2 × 1/2 × 0.450 × v²
238= 0.450v²
Divide both sides by 0.450
238/0.450 = 0.450v²/0.450
v² = 528.89
Square root both sides
Sq rt v² = sq rt 528.89
V = 22.998m/s
V is approximately = 23m/s
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Answer:
T1 = 130N, T2 = 370N
Explanation:
In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.
1. forces:
Let tension in rope 1 be T1 and in rope 2 be T2:
ma = T1 + T2 - 100N - 400N = 0
(1) T1 + T2 = 500N
2. torque around the center point of the beam:
τ = r x F = 5*T1 + 3*400N - 5*T2 = 0
(2) T1 - T2 = -240N
Solving both equations:
T1 = 130N
T2 = 370N