Anew event has been proposed for the winter olympics. an athlete will sprint 100 m, starting from rest, then leap onto a 20 kg bobsled. the person and bobsled will then slide down a 50-m-long ice-covered ramp, sloped at 20 degrees , and into a spring with a carefully calibrated spring constant of 2000 n/m. the athlete who compresses the spring the farthest wins the gold medal. lisa, whose mass is 40 kg, has been training for this event. she can reach a maximum speed of 12 m/s in the 100 m dash. how far x will lisa compress the spring?

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

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