Question

M A mass M-1.2 kg is released from rest. From the initial height of 75 cm above the ground, the mass drops from an free spring (neither compressed, nor extended) of constant k = 590 N/m. The maximum compression of the spring, due to the impact of the dropping mass is sh = 160 mm. Analyse the system with respect to the balance of energy, and the apply the conservation of energy, considering the ground level as reference of the calculation of the potential energy and calculate: The energy at selected points indicated in the table: Potential Kinetic Epl = Ex1 Elastic Belt int 1 J J Еро Et = Beli Point Top of o free spring) J J E = E21 = B2 Point 2 J J Calculate

Answer 1

I hi ch hal TV th= 160mm 1.2kg h = 0.16 m Principle question is Given i h = 75cm, h = 0.75m k= 590N/m. to be used while solving this is conservation of energy. According to which the total energy of a system remains constant with time if no external force on the system. The types of energies in given system Potential energyk Gravitational Spring is applied is Kinetic energy 9 colastic

can 2 toto الیا . Eg: Mgha + k k Cah)? To find ha, we apply CoE Conserve ation of energy) between pointi 1 and 2 Total at state 1 energy GPE, IKE, & EPE, Mgh, (since velocity to and compression (1.2) (918)(0.75) J of spring :o) E a 0.823 at state 2 a GPE + K € + EPE 2 Mgha +0+ KM) compression / clongation => [ (1.2)(9.8) ha) + (590)(0-16)? E,= Ez Uring Total energy TEN 2 LEPE is given by & kam?) where on= COE 8.82 11.76hat 7.552 k₂= 0.108m he height of free spring top surface =

ho ha + Dh ho ² (0.108+ 0.16) 3 0.268 m ho Point 1 Calculation of energies at different points Potential energy (gravitation) - Mghi (1-2)(9.8)(0.75) Epi 8.2 J Kinetic ene energy of Mys, since v=0 KE 'ki Elastic en energy ation zo ; since since spring E ell =/3.15 J = Epo Kinetic en 2 = £ kan? elongation Pointe: Potential energy Mg ho Epo = (12)69-8)(0.268) - 3 energy & MUZ velocity of block on reaching spring top It can be obtained by using 3rd equation v²_u²= 2a5 final relocity, initial velocity acceleration displacement ga = 9.8ms² sih-ho = 0.482 on rea of motion az عر Here V = Voj a- 2

v² = 21918) (0.482) = 3.074 m/s ..kę at point o Do & MV ² = 2 (1.2) (3.074) ² Eko = 5.67 J Ecco since spring's clongation compression to hz Point 2: Potential energy: Mg (12) 98 (0.108) : 1:27 J - Epz Epz 2 elastic potential energy Eel2 = å kan? Ź ( 590)(0.16) 7.552 J Eel,= 7.552 J Here, 2=0 since block will be at rest moment arily when the spring will be at manimum compression point 27 Ekz=0. KE at

6. At point 4 :- hy height abone ground = that th/3 = [0.100 m + 0.16 - - 0.161m hu= 0.161 m Mghu? . Epu= (12)(918)(0.161) = Epu = 1.89 J elongation of spring at point 422Ah 0.107m Eelt $(590)(0.107) 2 as Eel4 = 3.38 J + Feat Ex4 8.82 J z Eput Eu= E, - Eka: 882-(3-38 + 1-84) = 2) Ekg = 3.55J FM (Vu) 2 V4 2 / (3.55x2) 1.2 Vu= 2.43 m/s