Question

A mass m=88.6 kg slides on a frictionless track with initial velocity vA=12.3 m/s at Position A with height hA=42.4 m. It passes over a lower hill with a height hB=21.6 m (at Position B) before stopping by running into a large spring with spring constant k=5173 N/m at Position C at height hC=20.3 m. The mass is brought to a stop at Position D, after compressing the spring by a length of d. Find the speed of the object (vB) at position B in m/s and the amount the spring is compressed (d) in m. Select one value for the speed at position B (vB) and one for the distance (d).

Answer 1

Step-by-step explanation:

Speed of object at B :

loss of height = 42.4 - 21.6

= 20.8 m /s

loss of potential = mgh , m is mass , h is height lost

= 88.6 x 9.8 x 20.8

= 18060.22 J

= gain of kinetic energy

Initial kinetic energy = 1/2 x 88.6 x 12.3 ²

= 6702.14 J

Total kinetic energy at B

= 6702.14 + 18060.22

= 24762.36 J

1/2 m v² = 24762.36

.5 x 88.6 x v² = 24762.36

v² = 558.97

v = 23.64 m /s

energy at point C

= Kinetic energy at point B - loss of potential energy due to loss of height

= 24762.36 J - mgh ,

h is further loss of height = 21.6 - 20.3

= 1.3 m

24762.36 J - mgh

= 24762.36 J - 88.6 x 9.8 x 1.3

= 23633.6 J

this energy + loss of gravitational potential energy = gain of energy by spring

= 23633.6 J + mgd = 1/2 kd²

23633.6 J + 88.6 x 9.8 d = .5 x 5173x d²

2586.5 d² - 868.28 d - 23633.6=0

d² - .3357 d - 9.137=0

This is a quadratic equation which will give value of d .

d = 3.2 m

d =