A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N/m. The block is pulled to a distance x=10 cm from its equilibrium position at x=0 on a frictionless surface from rest at t=0. Calculate K.E ,P.E and T.E when it is 5 cm away from mean position.
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Answer:
PE=12.5,
KE=37.5
TE=50
Explanation:
50 n/m=0.5 N/cm
PE=-kx²
=-(0.5)(5)(5)
=12.5 joules
FOr ke
PE (initial)=-(0.5)(10)²
50J
according to conservation of energy
PE(initial)+KE(initial)=PE(final)+KE(final)
50+0=12.5+ke
KE=37.5
TE=KE+PE
=37.5+12.5
=50
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