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shrutij

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## Homework Statement

A horizontal spring with spring constant 92.2 N/m is compressed 15.5 cm and used to launch a 2.93 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.121. How far does the box slide across the rough surface before stopping?

## Homework Equations

KE=1/2mv^2

Usp=1/2kx^2

Work-energy theorem

## The Attempt at a Solution

The Kinetic energy gained by the box should be equal to the potential energy stored in the spring, so 1/2kx^2=1/2mv^2, substituting all values in, I isolated for the velocity, which gave me v=2.209 m/s. This is the launch velocity once the box leaves the spring.

Since it is a frictionless surface, this speed would be maintained (and becomes v0) when the box encounters the rough patch.

The work done by friction on the rough patch should thus be equal to the initial kinetic energy of the box. i.e. with speed 2.209 m/s, KEi = 7.15 J, and W by friction= Friction * d.

Using the coefficient of kinetic friction and the normal force, kinetic friction is 3.47 N, which gave me a d (how far the box traveled before coming to a stop) to be 2.06 m, which was wrong.

Where am I going wrong?