stuck YF14 14.P029 0/1 Submissions Used A 0.450-kg glider, attached to the end o

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YF14 14.P029 0/1 Submissions Used A 0.450-kg glider, attached to the end of an ideal spring with force constant k 475 N/m, undergoes SHM with an amplitude 0.040 m. (a) Compute the maximum speed of the glider m/s (b) Compute the speed of the glider when it is at x 0.015 m. m/s (c) Compute the magnitude of the maximum acceleration of the glider m/s2 (d) Compute the acceleration of the glider at x -0.015 m. m/s2 (e) Compute the total mechanical energy of the glider at any point in its motion.

Explanation / Answer

a)

when X = A, V = 0 so energy is

E = 1/2*k*A^2

E = 0.5 x 475 x 0.040^2

E = 0.38 J

when V = Vmax, X = 0

so 1/2*m*Vmax^2 = 0.38 J

Vmax^2 = (0.38 x 2) / 0.450

Vmax = 1.30 m/s

b)

1/2mv^2 + 1/2kx^2 = E

when x = -0.015 m

(0.5 x 0.450 x v^2) + (0.5 x 475 x (-0.015^2) = 0.38

v^2 = (0.38 + 0.0534) / 0.225

v = 1.39 m/s

c)

F = ma = -kx

max a occures when x is max, i.e x = +/- A

Imax a I = kA / m

Imax a I = (475 x 0.040) / 0.450

Imax a I = 42.22 m/s^2

d)

ma = -kx

a = -kx / m

a = (-475 x -0.015) / 0.450

a = 15.83 m/s^2

e)

see a) E = 0.38 J

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