a block with mass m is placed on a rough ramp with a 30° incline. the block remains at rest.
a. use the diagram on your student document to create a free body diagram for the block. label all forces on the block (not components).
note - the grid has been tilted to the same 30° angle as the incline for simplicity.
b. explain why the block remains at rest using your free - body diagram from part a.
c. the block is placed on a different incline that is also 30° but is so smooth that it is almost frictionless. describe the motion of the block on this new ramp
block does not move
constant velocity
constant acceleration
justify your claim using your free - body diagram from part a

Question

Accepted Answer

### Part A: Free - Body Diagram
The forces acting on the block are:
1. **Gravitational Force ($F_g = Mg$)**: Acts vertically downwards, where $M$ is the mass of the block and $g$ is the acceleration due to gravity.
2. **Normal Force ($F_N$)**: Acts perpendicular to the surface of the ramp (upwards and at an angle of $30^{\circ}$ from the vertical, or $60^{\circ}$ from the horizontal).
3. **Frictional Force ($F_f$)**: Acts parallel to the surface of the ramp, opposite to the direction of the component of the gravitational force that would cause the block to slide. Since the block is at rest, the frictional force is static friction and acts up the ramp.

To draw the free - body diagram:
- Draw a dot to represent the block.
- Draw an arrow pointing downwards to represent the gravitational force $F_g$.
- Draw an arrow perpendicular to the ramp (towards the block from the ramp) to represent the normal force $F_N$.
- Draw an arrow parallel to the ramp, pointing up the ramp, to represent the static frictional force $F_f$.

### Part B: Explanation for Rest
## Step 1: Resolve Gravitational Force
Resolve the gravitational force $F_g=Mg$ into two components: one parallel to the ramp ($F_{g\parallel}$) and one perpendicular to the ramp ($F_{g\perp}$). Using trigonometry, $F_{g\parallel}=Mg\sin(30^{\circ})$ and $F_{g\perp}=Mg\cos(30^{\circ})$.
## Step 2: Analyze Forces Parallel to Ramp
For an object at rest on an inclined plane, the net force parallel to the ramp is zero. The static frictional force $F_f$ opposes the component of the gravitational force that tries to pull the block down the ramp. So, $F_f = F_{g\parallel}=Mg\sin(30^{\circ})$. As long as the static frictional force is equal to the component of the gravitational force parallel to the ramp, the block will remain at rest (since $F_{net,\parallel}=F_f - F_{g\parallel}=0$). Also, the normal force $F_N$ balances the component of the gravitational force perpendicular to the ramp ($F_N = F_{g\perp}=Mg\cos(30^{\circ})$), so there is no net force perpendicular to the ramp ($F_{net,\perp}=F_N - F_{g\perp}=0$). Since the net force on the block $F_{net}=F_{net,\parallel}+F_{net,\perp} = 0$, according to Newton's first law of motion (an object at rest remains at rest if the net force acting on it is zero), the block remains at rest.

### Part C: Motion on Frictionless Ramp
## Step 1: Analyze Forces on Frictionless Ramp
On a frictionless ramp, the frictional force $F_f = 0$. The forces acting on the block are still the gravitational force $F_g = Mg$ (downwards) and the normal force $F_N$ (perpendicular to the ramp).
## Step 2: Resolve Forces and Find Net Force
Resolve the gravitational force into components parallel and perpendicular to the ramp. The component parallel to the ramp is $F_{g\parallel}=Mg\sin(30^{\circ})$, and the component perpendicular to the ramp is $F_{g\perp}=Mg\cos(30^{\circ})$. The normal force $F_N$ balances the perpendicular component ($F_N=F_{g\perp}$). The net force on the block is equal to the component of the gravitational force parallel to the ramp ($F_{net}=F_{g\parallel}=Mg\sin(30^{\circ})$) because there is no frictional force to oppose it.
## Step 3: Determine Motion
According to Newton's second law of motion $F = ma$, where $F$ is the net force, $m$ is the mass, and $a$ is the acceleration. Since $F_{net}=Mg\sin(30^{\circ})$ and $m = M$, we have $Mg\sin(30^{\circ})=Ma$. Canceling out the mass $M$ from both sides, we get $a = g\sin(30^{\circ})=\frac{g}{2}$ (assuming $g = 9.8\space m/s^2$, $a=4.9\space m/s^2$). The acceleration is constant (because the net force is constant, as the component of the gravitational force parallel to the ramp is constant for a given angle of inclination). So the block will move down the ramp with a constant acceleration. So the block does not move: No; Constant Velocity: No; Constant Acceleration: Yes.

### Justification for Part C
From the free - body diagram in part A, when the ramp is frictionless, the frictional force is zero. The only force component parallel to the ramp is the component of the gravitational force $F_{g\parallel}=Mg\sin(30^{\circ})$. According to Newton's second law $F = ma$, since there is a non - zero net force parallel to the ramp, the block will accelerate. The acceleration $a=\frac{F_{net}}{m}=\frac{Mg\sin(30^{\circ})}{M}=g\sin(30^{\circ})$, which is a constant value (because $g$ and $\sin(30^{\circ})$ are constant). So the block will move down the ramp with a constant acceleration.

### Answers:
- Part A: Free - body diagram with $F_g$ (down), $F_N$ (perpendicular to ramp up), $F_f$ (up ramp) as described.
- Part B: The block remains at rest because the net force on it is zero (static friction balances the parallel component of gravity, normal force balances the perpendicular component).
- Part C: Block does not move: $\boldsymbol{\text{No}}$; Constant Velocity: $\boldsymbol{\text{No}}$; Constant Acceleration: $\boldsymbol{\text{Yes}}$; Justification: Net force (parallel to ramp) is constant ($F_{net}=Mg\sin30^{\circ}$), so acceleration $a = g\sin30^{\circ}$ (constant) by $F = ma$.

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QUESTION IMAGE

Question

Explanation:

Part A: Free - Body Diagram

Part B: Explanation for Rest

Step 1: Resolve Gravitational Force

Step 2: Analyze Forces Parallel to Ramp

Part C: Motion on Frictionless Ramp

Step 1: Analyze Forces on Frictionless Ramp

Step 2: Resolve Forces and Find Net Force

Step 3: Determine Motion

Answer: