A 2.00 kg package is released on a53.1 o incline, 4.00 m from a long spring with
ID: 1758379 • Letter: A
Question
A 2.00 kg package is released on a53.1o incline, 4.00 m from a long spring with forceconstant 120 N/m that is attached at the bottom of the incline. Thecoefficients of friction between the package and the incline areµs=0.40 and µk=0.20. The mass ofthe spring is negligible.
a. What is the speed of thepackage just before it reaches the spring?
b. What is the maximum compression ofthe spring?
c. The package rebounds back upthe incline. How close does it get to its original position?
Explanation / Answer
(a) Fromthe diagram : velocity is : v = 2 g L ( sin - k cos) = ( 2 * 9.8 * 4.0 m ( sin53.1 - 0.20 cos53.1 ) ) = 7.30 m /s (b) This does require energy considerations; the combined work done bygravity and friction is : m g ( L + d) ( sin - k cos ) and the potential energy of the spring is : 1/2 kd2 where d is the maximum compression of the spring. This isa quadratic in d, which can be written as : d2 * [ k / 2 mg (sin - k cos) ] - d - L = 0 Simplifying the above we get d = 1.06 m (c) The easy thing to do here is to recognize that the presence of thespring determines d, but at the end of the motion thespring has no potential energy, and the distance below the startingpoint is determined solely by how much energy has been lost tofriction. If the block ends up a distance y below thestarting point, then the block has moved a distance L +d down the inclined and L + d - y up theinclined. y = ( L + d) * [2 k cos / ( sin + k cos ) ] = ( L + d) * [ 2k / ( tan + k ) ] = ( 4.00 + 1.06 ) * [ 2 * 0.20 / ( tan53.1 +0.20 )] simplifying y = 1.32 m Hope this helps u!Related Questions
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