Question

A 10 kg block on a horizontal surface is attached to a horizontal spring of spring constant k = 4.4 kN/m. The block is pulled to the right so that the spring is stretched 5.8 cm beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of 38 N. (a) What is the kinetic energy of the block when it has moved 2.2 cm from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

Answer 1

at point x1 = 5.8 cm = 0.058 m

energy stored E1 = (1/2)*k*X1^2

at point x2 = 2.2 cm = 0.022 m

at p

Energy E2 = (1/2)*m*v2^2 + (1/2)*k*x2^2

work done by friction Wf = -f*(x1-x2)

from energy conservation

E1 + Wf = E2

(1/2)*k*x1^2 - f*(x1-x2) = (1/2)*k*x2^2 + (1/2)*m*v2^2

(1/2)*k*(x1^2-x2^2) - f*(x1-x2) = (1/2)*m*v^2

(1/2)*4.4*10^3*(0.058^2-0.022^2) - 38*(0.058-0.022) = (1/2)*10*v^2

v = 0.997 m/s

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part b

at point x1 = 5.8 cm = 0.058 m

energy stored E1 = (1/2)*k*X1^2

at point x2 = 0

at p

Energy E2 = (1/2)*m*v2^2

work done by friction Wf = -f*(x1-x2)

from energy conservation

E1 + Wf = E2

(1/2)*k*x1^2 - f*(x1-x2) = K2

(1/2)*k*(x1^2-x2^2) - f*(x1-x2) = K2

(1/2)*4.4*10^3*(0.058^2-0^2) - 38*(0.058-0) = K2

K2 = 5.2 J

at point x the KE is maximum

K = (1/2)*k*(x1^2-x^2) - f*(x1-x)

dK/dx = 0

0 = -(1/2)*k*2x -f*(0-1)

0 = -(1/2)*k*2x + f

0 = -k*x + f

x = f/k = 38/(4.4*10^3) = 0.0086 m

maximum kinetic energy Kmax = (1/2)*k*(x1^2-x^2) - f*(x1-x)

maximum kinetic energy Kmax = (1/2)*4.4*10^3*(0.0581^2-0.0086^2) - 38*(0.058-0.0086) = 5.4 J