A 250 g block is dropped onto a vertical spring with springconstant k = 2.5 N/cm

Question

A 250 g block is dropped onto a vertical spring with springconstant k = 2.5 N/cm . The block becomes attatched tothe spring, and the spring compresses 12 cm before momentarilystopping. While the spring is being compressed, what work isbeing done on the block ( a) by its weight, and (b) by the springforce? (c) and What is the speed of the block just before it hitsthe spring? (assume that friction is negligible) (d) If thespeed at impact is doubled, what is the maximum compression of thespring?             Please show your final answers and how you arrived at each of them.Thank you.

Explanation / Answer

Given that the mass of block is m = 250 g = 0.250 kg          Thespring constant is K = 2.5 N/cm = 250 N/m           Thecompression in the spring is x = 12 cm = 0.12 m -------------------------------------------------------------------------       (a) The work done by thegravitational force is Wg = mgx                                                                                    =0.250kg*9.8m/s2*0.12m                                                                                    =--------- J      (b) work done by the soring isWs = - (1/2)kx2                                                             = --------- J      (c) From work energytheorem                                       -(1/2)mv2= Wg+Ws                                                  v = 2(Wg+Ws)/m     (d) If the velocity is doubled then                    U = 2v           Applyconservation of energy before and after the spring compressed                     totalenergy when spring is at initial length = totalenergy when spring is at compressed                     0 - (1/2)mu2 = mgd -(1/2)Kd2                                                                                           u2 =  2gd - Kd2 /m                                                 Then we get d = -------- m          Thespring constant is K = 2.5 N/cm = 250 N/m           Thecompression in the spring is x = 12 cm = 0.12 m -------------------------------------------------------------------------       (a) The work done by thegravitational force is Wg = mgx                                                                                    =0.250kg*9.8m/s2*0.12m                                                                                    =--------- J      (b) work done by the soring isWs = - (1/2)kx2                                                             = --------- J      (c) From work energytheorem                                       -(1/2)mv2= Wg+Ws                                                  v = 2(Wg+Ws)/m     (d) If the velocity is doubled then                    U = 2v           Applyconservation of energy before and after the spring compressed                     totalenergy when spring is at initial length = totalenergy when spring is at compressed                     0 - (1/2)mu2 = mgd -(1/2)Kd2                                                                                           u2 =  2gd - Kd2 /m                                                 Then we get d = -------- m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.