Question

Two horizontal forces, F and F₂, act on a box, but only Ę appears in the drawing. Ę₂ can refer to either the right or the left. The square moves along the x axis only. There is no friction between the box and the surface. Suppose F₁ = +4.0 N and the mass of the box is 4.1 kg. Find the magnitude and direction of F₂ when the acceleration of the box is (a) +7.0 m/s², (b) -7.0 m/s², and (c) 0 m/s².

Answer 1

Since there is no friction, the net force acting on the box is equal to the sum of the two horizontal forces. From Newton's second law, we know that the net force is equal to the mass of the box times its acceleration. Therefore:

ΣF = m * a

where ΣF is the net force, m is the mass of the box, and a is the acceleration of the box.

We can use this equation to find the magnitude of F₂ in each case.

(a) When the acceleration of the box is +7.0 m/s²:

ΣF = F₁ + F₂

m * a = F₁ + F₂

(4.1 kg) * (7.0 m/s²) = 4.0 N + F₂

F₂ = (4.1 kg) * (7.0 m/s²) - 4.0 N

F₂ = 25.7 N to the right

So, the magnitude of F₂ is 25.7 N, and it acts to the right.

(b) When the acceleration of the box is -7.0 m/s²:

ΣF = F₁ + F₂

m * a = F₁ + F₂

(4.1 kg) * (-7.0 m/s²) = 4.0 N + F₂

F₂ = (4.1 kg) * (-7.0 m/s²) - 4.0 N

F₂ = -32.6 N to the left

So, the magnitude of F₂ is 32.6 N, and it acts to the left.

(c) When the acceleration of the box is 0 m/s²:

ΣF = F₁ + F₂

m * a = F₁ + F₂

(4.1 kg) * (0 m/s²) = 4.0 N + F₂

F₂ = -4.0 N

So, the magnitude of F₂ is 4.0 N, and it acts to the left.

Step-by-step explanation: