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Answer:
(C) 40m/s
Explanation:
Given;
spring constant of the catapult, k = 10,000 N/m
compression of the spring, x = 0.5 m
mass of the launched object, m = 1.56 kg
Apply the principle of conservation of energy;
Elastic potential energy of the catapult = kinetic energy of the target launched.
¹/₂kx² = ¹/₂mv²
where;
v is the target's velocity as it leaves the catapult
kx² = mv²
v² = kx² / m
v² = (10000 x 0.5²) / (1.56)
v² = 1602.56
v = √1602.56
v = 40.03 m/s
v ≅ 40 m/s
Therefore, the target's velocity as it leaves the spring is 40 m/s
Answer:
a) 264.74 N
b) 91.15 N
c) 20.12°
Explanation:
Given:
Angle between the rope and the vertical, θ = 19°
Tension in the rope, T = 280 N
For the system to be in equilibrium,
The net force in vertical as well as in horizontal direction, should be zero
Therefore,
a) For Vertical direction
Weight of the child = vertical component of the tension
W = T cosθ ..............(1)
or
W = 280 cos19° = 264.74 N ............(a)
b) For horizontal
force on the child, F = T sinθ .............(2)
or
F = 280 sin19° = 91.15 N
c) Now, on dividing (1) and (2), we have
W/F = Tcosθ/Tsinθ
or
tanθ = F/W
now, for F = 97 N
tanθ = 97/264.74 = 0.3663 (W from (a))
or
θ = tan⁻¹(0.3663)
or
θ = 20.12°
