Question
PLEASE SOLVE CLEARLY , I NEED HELP !!
At right, a small object (mass = 100 kg) starts from rest at point (1) and slides down a curved vertical track until it reaches a horizontal segment between the points (2) and (3). We note that there is a spring (k = 96000 N/m) on the horizontal segment; the spring is initially in equilibrium. We can assume that the entire system is frictionless and that there is no air resistance. If the heights H = 360 m and h = 120 m then what will be the speed of the object after compressing the spring 0.50 m from equilibrium? (in m/sec)
Explanation / Answer
speed of block just before it touches spring ,
mv^2 /2 = mg (H - h)
v =sqrt(2 x 9.8 x (360 - 120)) = 68.59 m/s
after that
m 68.59^2 /2 = 96000* 0.5^2 /2 + mv^2 /2
100 x 68.59^2 /2 = 12000 + 100v^2 /2
v = 66.82 m/s