Question

# A 1.40 kg block slides with a speed of 0.950 m/s on a frictionless horizontal surface...

A 1.40 kg block slides with a speed of 0.950 m/s on a frictionless horizontal surface until it encounters a spring with a force constant of 734 N/m. The block comes to rest after compressing the spring 4.15 cm. Find the spring potential, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of (a) 0 cm, (b) 1.00 cm, (c) 2.00 cm, (d) 3.00 cm, (e) 4.00 cm

mass of the block=m=1.4 kg

initial speed=v=0.95 m/s

spring constant=k=734 N/m

compression of the spring=x=4.15 cm=0.0415 m

then using the conservation of energy principle

initial kinetic energy of the block=final potential energy of the spring

==>0.5*m*v^2=0.5*k*x^2

potential energy of the spring=0.5*k*x^2=0.5*734*0.0415^2=0.6321 J

kinetic energy oof the block =0.5*mass*speed^2=0.5*1.4*0.95^2=0.63175 J

total mechanical energy of the system=0.6321 J (higher of the two values)(assuming the values given have some inherent errors due to rounding off/significant number due to which the energy values are closer but not equal)

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