Question

force Fx acts on a particle that has a mass of 2 kg. The force is related to the position x of the particle by the formula F 3, where C-0.5 if x is in meters and Fy is in Newton (a)What are the SI units of C? (Use the following as necessary: N and m.) (b) Find the work Wdone by this force as the partidle moves from x - 2.8 m to x - 1.6 m (C) At x= 1.6 m the force points opposite the direction of the particle's velocity (speed is 12 m/s), what is its speed at 2.8 m? m/s Can you tell its direction of motion at x-2.8 m using only the work-kinetic energy theorem yes O no Explain. has not been graded yet

Answer 1

a)

F_x = C x³

N = C [m³]

C = Nm^-3

so unit are Newton/meter³

b)

work done is given as

W = $\int_{x_{i}}^{x_{f}}$ F dx

given that x_i = 2.8 , x_f = 1.6 m

W = $\int_{2.8}^{1.6}$ C x³ dx

W = (C/4) (1.6⁴ - 2.8⁴)

W = (0.5/4) (1.6⁴ - 2.8⁴)

W = - 6.86 J

c)

v_i = initial speed at 1.6 m, = 12 m/s

v_f = final speed = ?

using work-change in kinetic energy

W = (0.5) m (v_f² - v_i²)

6.86 = (0.5) (2) (v_f² - (12)²)

v_f = 12.3 m/s