Question

4.4 q: which diagram represents a box in equilibrium? 4.5 q: if the sum of all the forces acting on a moving object is zero, the object will (a) slow down and stop (b) change the direction of its motion (c) accelerate uniformly (d) continue moving with constant velocity 4.9 q: what is the weight of a 2.00 - kilogram object on the surface of earth? (a) 4.91 n (b) 2.00 n (c) 9.81 n (d) 19.6 n

Explanation:

4.4 Q:

Step1: Recall equilibrium condition

For an object to be in equilibrium, the net - force acting on it in all directions must be zero. In the case of a two - dimensional force diagram (horizontal and vertical), the sum of horizontal forces $\sum F_x = 0$ and the sum of vertical forces $\sum F_y=0$.

Step2: Analyze each diagram

In diagram (1), horizontally the forces are equal ($5N$ to the left and $5N$ to the right), and vertically the forces are equal ($3N$ up and $3N$ down). So, the net force in both directions is zero.
In diagram (2), horizontally the forces are equal ($2N$ to the left and $2N$ to the right), but vertically $8N$ down and $2N$ up, so $\sum F_y
eq0$.
In diagram (3), horizontally $8N$ to the left and $2N$ to the right, so $\sum F_x
eq0$, and vertically $2N$ up and $2N$ down.
In diagram (4), horizontally $5N$ to the left and $3N$ to the right, so $\sum F_x
eq0$, and vertically $2N$ up and no other vertical force mentioned.
The box in equilibrium is represented by diagram (1).

4.5 Q:

Step1: Apply Newton's first law

According to Newton's first law of motion, if the net force $\sum F = 0$ acting on a moving object, the object will continue to move with a constant velocity because there is no unbalanced force to change its state of motion. A non - zero net force is required to slow down, change direction, or accelerate an object.
So the object will continue moving with constant velocity.

4.9 Q:

Step1: Use weight formula

The weight $W$ of an object on the surface of the Earth is given by the formula $W = mg$, where $m$ is the mass of the object and $g$ is the acceleration due to gravity on the surface of the Earth ($g = 9.8m/s^{2}$).
Given $m = 2.00kg$, then $W=mg=2.00kg\times9.8m/s^{2}=19.6N$.

Answer:

4.4 Q: Diagram (1)
4.5 Q: (D) continue moving with constant velocity
4.9 Q: (D) 19.6 N