Question

A 3.2-kg object is subjected to two constant forces. F_1 = (1.9 N)i + (-1.9 N)j and F_2= (3.8 N)i +(-10.1 N)j. The object is at rest at origin at lime t = 0.

A) What is the object's acceleration?
B) What is its velocity at time t = 3.0 s?
C) Where is the object at time t = 3.0 s?

Answer 1

A)

a= 4.15 m/s² : Magnitude of the acceleration

β= 64.59° : Direction of the acceleration with respect to the horizontal

B)

v = 12.45 m/s : Magnitude of the velocity

β= 64.59°: Direction of the velocity with respect to the horizontal

C)

d = 18.675 m : Magnitude of the displacement

β= 64.59° : Direction of the displacement with respect to the horizontal

Explanation:

A)Data

m=3.2 kg

F₁= (1.9 N)i + (-1.9 N)j

F₂= (3.8 N)i +(-10.1 N)j.

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

We apply the (Formula 1)

F₁ + F₂= m*a

( (1.9 N)i + (-1.9 N)j)+((3.8 N)i +(-10.1 N)j) = 3.2* a

5.7 i - 12 j = 3.2* a

a=( 5.7/ 3.2) i - (12/3.2) j

`a=\sqrt{( 5.7/ 3.2) ^(2) + (12/3.2) ^(2)  }`

a= 4.15 m/s² : Magnitude of the acceleration

`\beta = tan^(-1) ((12/3.2)/(5,7/3.2) )`

`\beta = tan^(-1) ((12)/(5,7) )`

β= 64.59°: Direction of the acceleration with respect to the horizontal

B)-C) Because the object moves with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

a= 4.15 m/s²

t = 0,

t = 3.0 s

B) we apply the formula (2)

vf= v₀+at

vf= 0+(4.15)(3)

vf= 12.45 m/s :magnitude of the velocity

β= 64.59°: Direction of the velocity with respect to the horizontal

C) we apply the formula (3)

d= v₀t+ (1/2)*a*t²

d= 0+ (1/2)(4.15)(3)²

d=18.675 m :Magnitude of the displacement

β= 64.59°: Direction of the displacement with respect to the horizontal