Question

An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces. F1 is 28.6 N acting due East, F2 is 32.8 N acting due North, F3 is 47.8 N acting due West, and F4 is 17.4 N acting due South. How much total work is done on the object in 2.22 seconds, if it has a mass of 20.0 kg?

Which type of energy is changing for the object while the above work is being done?

How fast does the object end up moving at the end of the 2.22 seconds?

Answer 1

Net northery force, F(n) = F2 - F4 = 32.8 - 17.4 = 15.4N
Net westerly force, F(w) = F3 - F1 = 47.8 - 28.6 = 19.2 N
Resultant force, F(r) = (F(n)² + F(w)²)^1/2 = sqrt(15.4² + 19.2²) = 24.61 N
Acceleration, a = F(r) / m = 24.61 N / 20.0 kg = 1.23 m/s²
Distance traveled in 2.22 s, d = (1/2) * a * t²
d = (1/2) * 1.23m/s² * (2.22s)² = 3.03 m
Work, W = F * d = 24.61 N * 3.03 m = 74.56 Nm

V = a * t = 1.23 m/s² * 2.22 s = 2.73 m/s

potential energy is changing into kinetic energy in this process.