(7%) constant k = 720 N/m. It is initially at rest on an inclined plane that is
ID: 1633260 • Letter: #
Question
(7%) constant k = 720 N/m. It is initially at rest on an inclined plane that is at an angle of = 29° with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is ,-0.11. In the initial position, where the spring is compressed by a distance of d= 0.11 m,the mass is at its lowest position and the spring is compressed the maximum amount. Take the initial gravitational energy of the block as zero. 2: A block of mass 2.5 kg is attached to a spring with spring i d l 0 Randomized Variables m 2.5 kg k = 720 N/m d=0.11 m Pk=0.11 -q) 50% Part (a) what is the block's initial mechanical energy. EO in J? spring remains attached to both the block and the fixed wall throughout its motion. ©theexpertta.com 50% Part (b) If the spring pushes the block up the incline, what distance L in meters, will the block tra el before coming to rest? The L= Potential 88% emaining: 2 attempt) cotanO asin0 acos0Explanation / Answer
b) let's use work energy theorem to find the distance travlled (L), before coming to rest
Work done= Gravtational PE + Elastic PE
Coefficient of friction ( force applied ) ( displacement) = mgLsin 29 + 1/2 K ( d^2-L^2)
plugging the values,
0.11 ( 2.5 ) (9.8) ( sin 29 ) ( L) = 2.5 (9.8) (L) sin 29 + 1/2 (720 ) ( 0.11^2- L^2)
1.3065 L - 11.877 L = -360 L^2 + 4.356
360L^2 - 10.57 L - 4.356=0
solving for L,
L = ( 10.57 +- 79.902) / 720) = 0.125 m apprx
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