\begin{tabular}{lll} 3.1 State Newton's second law in words. \ 3.2 & Draw a free body diagram of ALL forces acting on the block. \ 3.3 & $$ \begin{array}{l} ext { Calculate the magnitude of: } \ ext { 3.3.1 The vertical component of the applied force. } \ ext { 3.3.2 The normal force. }\end{array} $$ \ $$ \begin{array}{ll} ext { 3.3.3 Frictional force. }\end{array} $$ \ 3.4 & $$ \begin{array}{l} ext { What will happen to the coefficient of frictional force between the block } \ ext { and the surface if the angle of the applied force is decreases from } 30^{\circ} \ ext { to } 26^{\circ} ext { ? } \ ext { Write down only INCREASES, DECREASES or REMAINS THE SAME. }\end{array} $$ \ 3.5 & $$ \begin{array}{l} ext { (2) }\end{array} $$ \ $$ \begin{array}{l} ext { The block above is now removed from the horizontal surface and then } \ ext { Placed on a slope that makes an angle of } 30^{\circ} ext { with the horizontal as } \ ext { shown in the diagram below. }\end{array} $$ \ \hline\end{tabular}

Question

\begin{tabular}{lll} 3.1 State Newton's second law in words. \ 3.2 & Draw a free body diagram of ALL forces acting on the block. \ 3.3 & $$ \begin{array}{l}	ext { Calculate the magnitude of: } \ 	ext { 3.3.1 The vertical component of the applied force. } \ 	ext { 3.3.2 The normal force. }\end{array} $$ \ $$ \begin{array}{ll}	ext { 3.3.3 Frictional force. }\end{array} $$ \ 3.4 & $$ \begin{array}{l}	ext { What will happen to the coefficient of frictional force between the block } \ 	ext { and the surface if the angle of the applied force is decreases from } 30^{\circ} \ 	ext { to } 26^{\circ} 	ext { ? } \ 	ext { Write down only INCREASES, DECREASES or REMAINS THE SAME. }\end{array} $$ \ 3.5 & $$ \begin{array}{l}	ext { (2) }\end{array} $$ \ $$ \begin{array}{l}	ext { The block above is now removed from the horizontal surface and then } \ 	ext { Placed on a slope that makes an angle of } 30^{\circ} 	ext { with the horizontal as } \ 	ext { shown in the diagram below. }\end{array} $$ \ \hline\end{tabular}

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### 3.1 State Newton's second law in words.
Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In simpler terms, the greater the force applied to an object, the greater its acceleration, and the more massive the object, the less it will accelerate in response to the same force.

### 3.2 Draw a free body diagram of ALL forces acting on the block.
In a free body diagram for a block on a horizontal surface with an applied force at an angle, the following forces should be represented:
- The weight of the block ($$W$$) acting downwards.
- The normal force ($$N$$) acting upwards.
- The applied force ($$F$$) at an angle, which can be broken down into two components: the vertical component ($$F_v$$) and the horizontal component ($$F_h$$).
- The frictional force ($$f$$) acting opposite to the direction of motion.

### 3.3 Calculate the magnitude of:
#### 3.3.1 The vertical component of the applied force.
The vertical component of the applied force can be calculated using the formula:
$$
F_v = F \sin(\theta)
$$
where $$F$$ is the magnitude of the applied force and $$\theta$$ is the angle of the applied force.

#### 3.3.2 The normal force.
The normal force can be calculated using the equation:
$$
N = W - F_v
$$
where $$W$$ is the weight of the block, and $$F_v$$ is the vertical component of the applied force calculated in the previous step.

#### 3.3.3 Frictional force.
The frictional force can be calculated using the formula:
$$
f = \mu N
$$
where $$\mu$$ is the coefficient of friction and $$N$$ is the normal force.

### 3.4 What will happen to the coefficient of frictional force between the block and the surface if the angle of the applied force decreases from $$30^{\circ}$$ to $$26^{\circ}$$?
The coefficient of friction ($$\mu$$) is a property of the surfaces in contact and does not change with the angle of the applied force. Therefore, the answer is:
**REMAINS THE SAME.**

### 3.5 The block above is now removed from the horizontal surface and then placed on a slope that makes an angle of $$30^{\circ}$$ with the horizontal.
In this case, the forces acting on the block will change due to the incline. The weight of the block will have a component acting down the slope, and the normal force will be perpendicular to the slope. The frictional force will also depend on the normal force and the coefficient of friction.

If you have specific values for the applied force, weight of the block, or coefficient of friction, please provide them so we can perform the calculations for parts 3.3.1, 3.3.2, and 3.3.3.
**3.1** Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. **3.2** Draw a free body diagram showing: - Weight ($$W$$) acting downwards. - Normal force ($$N$$) acting upwards. - Applied force ($$F$$) at an angle, split into vertical ($$F_v$$) and horizontal ($$F_h$$) components. - Frictional force ($$f$$) opposing motion. **3.3** - **3.3.1** Vertical component of applied force: $$F_v = F \sin(\theta)$$ - **3.3.2** Normal force: $$N = W - F_v$$ - **3.3.3** Frictional force: $$f = \mu N$$ **3.4** The coefficient of friction remains the same when the angle of the applied force decreases from $$30^{\circ}$$ to $$26^{\circ}$$. **3.5** When the block is placed on a $$30^{\circ}$$ slope, the forces adjust accordingly, with weight and normal force components along the slope.

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