A cart w 050 kg is attached to an odd spring The force exer

A cart (w = 0.50 kg) is attached to an odd spring. The force exerted by the spring is given by F_a = -bx^2, where b = 30 N/m^2. Compute the work done by the spring\'s force as the cart moves from x_1 to x_2. If the cart is released from rest at x_0 = -15 cm, what is its speed when x = 0?

Solution

a) Work done = intgeral (F.dx)

= integral ( -bx^2) dx from x1 to x2

= - b(x^3/3) from x1 to x2

= -30/3 ( x2^3 - x1^3)

= 10(x1^3 - x2^3)

b) work done = 10(0.15^3 - 0 )

= 0.03375 J

work done = change in KE

w = 0.5mv^2

0.03375 = 0.5*0.5*v^2

v = 0.3674 m/s