Question

1. A person weights 120 lb. Determine

(a) his mass in kilograms.

(b) his weight in newtons.

2. Three forces acting on an object are given by F₁ = (–2.00i + 2.00j) N, F₂ = (5.00i – 3.00j) N, and F₃ = (–45.0i) N. The object experiences an acceleration of magnitude 3.75 m/s2 .

(a) What is the direction of the acceleration?

(b) What is the mass of the object?

(c) If the object is initially at rest, what is its speed after 10.0 s?

(d) How far has it moved in this 10.0 s?

Answer 1

1.(a) 1 lb = 0.453 kg

hence m = 120 lb = 54.43 kg

(b) wight = mg = 54.43*9.8 = 533.41 N

2. The Net force F = F1+F2+F3   (all are vectors, hence vector addition)

F = (-2.00+5.00-45.00) i + (2.00-3.00) j

= -42.00 i -1.00 j

which is the direction of acceleration same as the direction of force.

(b) the magnitude of force

hence mass = force/acceleration = 42.01/3.75 = 11.20 kg

(c) initial velocity u = 0

t = 10.00 sec

acceleration = a = 3.75

then final velocity v = u+at = at = 3.75*10.00 = 37.5 m/sec²

(d) the distance s= ut + (1/2) at² = 0.5*3.75*100 = 187.5 m