Question

A block with mass m = 7.3 kg is attached to two springs with spring constants k_left = 30 N/m and k_right = 52 N/m. The block is pulled a distance x = 0.25 m to the left of its equilibrium position and released from rest.

Where is the block located, relative to equilibrium, at a time 1.03 s after it is released? (if the block is left of equilibrium give the answer as a negative value; if the block is right of equilibrium give the answer as a positive value)

What is the net force on the block at this time 1.03 s? (a negative force is to the left; a positive force is to the right)

Answer 1

given

m = 7.3 kg

xi = -0.25 m

k1 = 30 N/m

k2 = 52 N/m

at t = 1.03, x = ?

effective spring constant, k = k1 + k2

= 30 + 52

= 82 N/m

angular frequency of motion, w = sqrt(k/m)

= sqrt(82/7.3)

= 3.35 rad/s

amplitude of motion, A = 0.25 m

so, the equation for the position of the block,

x = -A*cos(w*t)

= -0.25*cos(3.35*t)

at time t = 1.03 s,

x = -0.25*cos(3.35*1.03)

= 0.238 m <<<<<<<<<<<------------Answer

Net force, Fnet = -k*x

= -82*0.24

= 19.7 N <<<<<<<<<<<------------Answer