Question

A mass m=29.0 kg slides on a frictionless track with initial velocity v_A=16.5 m/s at Position A with height h_A=53.1 m. It passes over a lower hill with a height h_B=26.4 m (at Position B) before stopping by running into a large spring with spring constant k=5058 N/m at Position C at height h_C=23.5 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 (v_B) 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

(a) Mass, m = 29.0 kg

At Position A -

Velocity, vA = 16.5 m/s

hA = 53.1 m

At position B -

Velocity, vB = ?

hB = 26.4 m

Now,

Total energy of mass at A = Total energy of mass at B

=> (1/2)*m*vA^2 + m*g*hA = (1/2)*m*vB^2 + m*g*hB

=> 0.5*vA^2 + g*hA = 0.5*vB^2 + g*hB

=> 0.5*16.5^2 + 9.8*53.1 = 0.5*vB^2 + 9.8*26.4

=> 136.13 + 520.38 = 0.5*vB^2 + 258.72

=> 0.5*vB^2 = 397.79

=> vB^2 = 795.58

=> vB = 28.2 m/s (Answer)

(b) Consider the position C -

Total energy of mass at A = Total energy of mass at C

=> (1/2)*m*vA^2 + m*g*hA = (1/2)*m*vC^2 + m*g*hC

=> 0.5*vA^2 + g*hA = 0.5*vC^2 + g*hC

=> 0.5*16.5^2 + 9.8*53.1 = 0.5*vC^2 + 9.8*23.5

=> 136.13 + 520.38 = 0.5*vC^2 + 230.3

=> 0.5*vC^2 = 426.21

=> vC^2 = 852.42

=> vC = 29.2 m/s

At position D, all the energy will be stored in the spring as spring energy.

Means -

(1/2)*m*vC^2 = (1/2)*k*d^2

=> 0.5*29*852.42 = 0.5*5058*d^2

=> d^2 = (29*852.42) / 5058

=> d = 2.21 m (Answer)