2. A 200-g block is launched by compressing a spring of constant k = 150 N/m by

Question

2. A 200-g block is launched by compressing a spring of constant k = 150 N/m by 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. Beyond the equilibrium position of the spring end, the surface has frictional coefficient k = 0.25. This frictional surface extends 85 cm. followed by a frictionless curved rise. Where does the block finally come to rest? Measure from the left end of the frictional zone. Note: The block can make multiple trips over the patch with friction. frictionless frictionless

Explanation / Answer

Energy stored in the spring by compressing it is -

E = (1/2)*k*x^2 = 0.5 * 150 * 0.15^2 = 1.6875 J

Now this energy shall be utilized in the work done against the friction.

Frictional force, F = mu*m*g = 0.25*0.20*9.81 = 0.4905 N

Suppose the block travels a distance d meter on this frictional surface.

then we have -

F*d = 1.6875

=> 0.4905*d = 1.6875

=> d = 1.6875 / 0.4905 = 3.44 m

length of the frictional surface = 85 cm = 0.85 m

by completing two complete trips on this frictional surafce, distance covered by the block = 0.85 x 4 = 3.40 m

remaining distance = 3.44 - 3.40 = 0.04 m = 4.0 cm

So, the block will stop at a distance of 4 cm from its initial position at the frictional surface.

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