Work and Energy

A 0.62 kg wood block is firmly attached to a very light horizontal spring (k=180 N/m). it is noted that the block-spring system, when compressed 5cm and released, stretches out 2.3cm beyond the equilibrium position before stopping and turning back. what is the coefficient of kinetic friction between the block and the floor?

Easy Math Editor

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Comments

kindly answer please.. need some expert.. thanks a lot..

Jon-jon Castro - 7 years, 8 months ago

the potential energy stored in the spring block system at the initial stage will be 1/2kx^2 and this energy will be dissipated against the friction when its being released from the compressed configuration By energy conservation we get 1/2kx^2=U(kinetic) [since its not static} mg*(x+2.3) where x=5cm. solving we get u(kinetic)= 0.5073

Ramesh Goenka - 7 years, 8 months ago

thank you so much..

Jon-jon Castro - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$