Answer:
3.5 m
Explanation:
Given
Mass of the bob sled , m_b = 20 kg
Mass of the girl, m = 40 kg
Speed of the girl, v = 12 m/s
Spring constant , k = 2000 N/m
Length of the ramp, L = 50 m
Angle of incline , θ = 20°
When the girl leaps on the sled , we use conservation of momentum principle
to find the speed of sled and the girl
m v + m_b (0) = (m + m_b) v_1
40 kg ×12 m/s = (20 kg + 40 kg ) v_1
v_1= 8 m/s
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The height of the incline is ,
h = L sin 20 = 50 m sin 20° = 17.1 m
By law of conservation of energy
(m + m_b) g h + 1/2 (m + m_b) v_1^2 = 1/2 (m +m_b) v2^2 + 1/2 k x^2
(60 kg ) ×9.8 m/s^2 ×17.1 m + 1/2 (60 kg ) (8 m/s)^2 = 1/2 (m +m_b)(0)^2 +1/2(2000 N/m) x^2
x = 3.5 m
Thus, the compression in the spring is 3.5 m
Lisa would compress the spring by an extension of 3.5 meters.
Given the following data:
First of all, we would determine the speed of the sled and the girl by applying the law of conservation of momentum:
[tex]m v + m_b (u) = (m + m_b) v_1\\\\m v + m_b (0) = (m + m_b) v_1\\\\40 \times 12 = (20 + 40 ) v_1\\\\480=60v_1\\\\\v_1= 8 m/s[/tex]
For height of the incline:
[tex]h = L sin 20 \\\\h= 50 \times sin 20[/tex]
h = 17.1 m.
From the law of conservation of energy, the potential and kinetic energy of the sled and girl is equal to the elastic and kinetic energy possessed by both the sled and girl and the spring.
[tex](m + m_b) g h + 1/2 (m + m_b) v_1^2 = \frac{1}{2} (m +m_b) v_2^2 + \frac{1}{2} k x^2\\\\((20+40 ) \times 9.8 \times17.1 ) + \frac{1}{2}\times (20+40 ) (8 )^2 = \frac{1}{2}\times (m +m_b)(0)^2 + \frac{1}{2}(2000 ) x^2[/tex]
Extension, x = 3.5 m.
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