One end of a horizontal spring with force constant 1300 Nm i

One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 4.00-kg block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is k = 0.400. You apply a constant force F to the block. F has magnitude 86.0 N and is directed toward the wall. The spring is compressed 80.0 cm.

What is the speed of the block?

What is the magnitude of the block\'s acceleration?

Solution

friction force, f = uk N = uk m g = 0.4 x 4 x 9.8 = 15.68 N

Now using work-energy theorem,

Work done by Force + work done by friction + work done by spring = change in KE

(0.80 * F) + ( - 0.80f) + ( - 130(0.80^2)/2) = m v^2 /2 - 0

68.8 - 12.544 - 41.6 = 4 v^2 / 2

v = 2.71 m/s ........Ans

at this time. Applying Fnet = m a

Fnet = F - kx - f = ma

86 - (130 x 0.8) - 15.68 = 4a

a = - 8.42 m/s^2    (minus indicates that in opposite direction of velocity )