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
