Question

Each of the two identical springs in the figure below has force constant k = 1.16 103 N/m, d = 1.21 m, and θ = 29.0°. (a) Find the unstretched length of each spring. (b) Find the instantaneous acceleration of the weight if it is pulled 10.0 cm lower and released.

Answer 1

a) If the system is in equilibrium, the net force is equal to zero.

2*F*sin(theta) - W = 0

then, the force is,

F = W/(2*sin(theta))= 100/(2*sin(29)) = 103.13 N

the restoring force is,

F = k*x

then, the compressed length is,

x = F/k =103.13/(1.16*10^3) = 0.0889 m

so, the length Lo is calculated as follows:

Lo = d-x = 1.21 - 0.0889 =1.1211 m = 1.12 m

b)

b)

the vertical distance from top is,

y = 0.1 + d*sin(29) = 0.1 + 1.21*sin(28.5) = 0.6866 m

the angle made by the spring with horizonatl is,

= tan^-1[ 0.6866/(1.21*cos(29))] = 32.97^o

the length of the springs is,

d' = 0.6866/sin(32.97) = 1.26 m

hence, the new compressed length is,

x' = d' - Lo = 1.26 - 1.12 = 0.14 m

hence, the force exrted by each spring is calculated as follows:

F = k*x'= 1.16*10^3*0.14 = 162.4 N

net force acting in upward direction is written as:

ma = 2*F*sin(32.97) - W

(W/g)*a = 2*F*sin(32.97) - W

hence, the acceleration of the block is,

a = 2*F*sin(32.97)*g/W - g = 2*162.4*sin(32.97)(9.8)/100 - 9.8 = 7.52 m/s²