In the figure show below, a 450 g box is placed in front of a spring, with sprin

Question

In the figure show below, a 450 g box is placed in front of a spring, with spring constant k, that is compressed a distance delta x = 18.1 cm from its natural length. When the box is let go, the spring will push the box and it will slide up the frictionless hill of height H = 1.9 m. At the top of hill, the ground is not frictionless. The coefficient of kinetic friction in this area is 0.20 and the coefficient of static friction is 0.30. The box will travel a distance d = 0.9 m on the rough surface at the top of the hill before coming to a stop. Determine the spring constant, k, of the spring.

Explanation / Answer

using work energy principle:

initial potential energy of the spring - work done against friction=final potential energy of the block

let spring constant be k N/m.

then initial potential energy of the spring=0.5*spring constant*compression^2

=0.5*k*0.181^2

friction force=kinetic friction coefficient*normal force

=0.2*mass*g

=0.2*0.45*9.8

=0.882 N

magnitude of work done against friction=friction force*distance

=0.882*0.9

=0.7938 J

final potential energy of the box=mass*g*height

=0.45*9.8*1.9

=8.379 J

using these values in the work -energy equation:

0.5*k*0.181^2-0.7938=8.379

==>k=(8.379+0.7938)/(0.5*0.181^2)

=559.98 N/m

so spring constant is 560 N/m. (approximately)

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