A block of mass 12.0 kg slides from rest down a frictionless 24.0° incline and i

Question

A block of mass 12.0 kg slides from rest down a frictionless 24.0° incline and is stopped by a strong spring with a spring constant 34.0x 103 N/m. The block slides 5.6 m from the point of release to the point where it comes to rest against the spring.

Use 10 N/kg for g.

a) What is in Joules the gravitational potential energy of the block the instant it is released, relative to the position where the block comes to a momentary rest against the spring?

b) What is in Joules the mechanical energy of the block-spring system?

c) How far in meters does the spring compress when the block comes to a momentarily stop?

Explanation / Answer

part A: Gravitational Potential energy U = mg(H) sin theta

U = 12 * 9.8 * 5.6 * sin 24

U = 267.86 Joules

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total Mech energy TE = EPE= U

TE = 267.86 J

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use 0.5 kx^2 = 267.86

x^2 = 2* 267.86/34e 3

x = 0.125 m or 12.5 cm

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