A 10 -Kg block on a horizontal surface is attached to a light spring (force cons

Question

A 10 -Kg block on a horizontal surface is attached to a light spring (force constant = 0.80 kN/m). The block is initially at rest at its equilibrium position when a force (magnitude P = 67.75 N) acting parallel to the surface is applied to the block, as shown. a) Assume the surface is frictionless. What is the speed of the block when it is 10 cm from its equilibrium position? b) If instead the coefficient of kinetic friction between the horizontal surface and the block was 0.15, what would the speed of the block be when it is 10cm from its equilibrium position?

Explanation / Answer

3) (A) Applying work - enegy theorem,

Work done by all forces = change in KE

Work done by force + work done by spring = change in KE

{ work done by gravity = work done by normal force = 0 }

(67.75 x 0.10) - (800 x 0.10^2 /2 ) = 10 (v^2 - 0) / 2

6.775 - 4 = 5 v^2

v = 0.745 m/s .........Ans

(b) N = m g

and f = uk m g = 0.15 x 10 x 9.81 = 14.7 N

now Applying work energy theorem,

work done by force + work done by spring + work done by friction = change in KE

6.775 - 4 - (14.7 x 0.10) = 5 v^2

v = 0.51 m/s

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