Question #167758

A spring is fixed to a wooden board with 2 masses suspend from it. The spring has stretched 6 cm. If m1 = 13 kg and m2 = 3 kg, what is the spring constant of the spring?


Expert's answer

Stretch in spring x=6cm=0.06mx=6cm=0.06m

Tw masses m1=13kg,m2=3kgm_1=13kg,m_2=3kg


Let k be the spring constant -


Then Restoration force in the spring F=kx=0.06k      −(1)F=kx=0.06k~~~~~~-(1)


Then Weight of masses = (m1+m2)g=(13+3)(9.8)(m_1+m_2)g=(13+3)(9.8)

=16×9.8=156.8N     −(2)=16\times 9.8=156.8N~~~~~-(2)


From eqn.(1) and (2) we have-

0.06k=156.8⇒k=156.80.06=2613.33N/m0.06k=156.8\Rightarrow k=\dfrac{156.8}{0.06}=2613.33N/m


hence The spring Constant is 2613.33N/m.


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