Question

A certain spring exerts a nonlinear force given by F(x) = −60x − 18x2 , where x is in meters and F is in newtons. A 0.90-kg block on a frictionless, horizontal surface is attached to the spring at x = 0. The block is pushed to x = −0.50 m and is then released. a) The work done by the spring on the block as it moves from x = −0.50 m to x = 0? b) The block’s speed when it reaches x = 0 c) What are the units of the coefficient with value –18?

Answer 1

a) W = 6.75 J and b) v = 3.87 m / s

Step-by-step explanation:

a) In the problem the force is nonlinear and they ask us for work, so we must use it's definition

W = ∫ F. dx

Bold indicates vectors. In a spring the force is applied in the direction of movement, whereby the scalar product is reduced to the ordinary product

W = ∫ F dx

We replace and integrate

W = ∫ (-60 x - 18 x²) dx

W = -60 x²/2 -18 x³/3

Let's evaluate between the integration limits, lower W = 0 for x = -0.50 m, to the upper limit W = W for x = 0 m

W = -30 [0- (-0.50) 2] -6 [0 - (- 0.50) 3]

W = 7.5 - 0.75

W = 6.75 J

b) Work is equal to the variation of kinetic energy

W = ΔK

W = ΔK = ½ m v² -0

v =√ 2W/m

v = √(2 6.75/ 0.90)

v = 3.87 m / s