I need the answer for B Part A Problem 10.72 What is the maximum compression of
ID: 1778099 • Letter: I
Question
I need the answer for B
Explanation / Answer
Given
mass of box m = 11 kg, sliding down the incline is 4.0 m , with angle of inclination is theta= 30 degrees
spring constant k = 230 N/m
Part A , maximum compression is Ds max = 1.62 m,
mg sin theta*xi = mg sin theta*xf + 0.5*k*x^2
11*9.8*4*sin30 = 11*9.8 *x*sin 30+0.5*230*x^2 ==> x = -1.62 m
Paart B
we know that the box will have max speed or max kientic energy at equilibrium position (wher the net force =0)
so the force at maximum displacement is
Fx = -kx -mg sin theta
equating Fx = 0
0 = -kx - mg sin theta
kx = mg sin theta
x = mg sin theta /k
x = 11*9.8 sin 30 /(230) m
x = 0.2343 m
so here x = -0.2343 m