Answer :
The block's resulting motion will be c) The block moves downward along the slope.
When a force is applied parallel to the slope, it can be resolved into two components: one perpendicular to the slope and one parallel to the slope. The force perpendicular to the slope counteracts the gravitational force acting on the block, while the force parallel to the slope overcomes frictional forces. In this case, the force exceeds 182 N, meaning it is sufficient to overcome friction, causing the block to move downward along the slope due to gravity.
To analyze the motion, we can use trigonometry to determine the components of the applied force. The force parallel to the slope is given by [tex]\( F_{\text{parallel}} = F \sin(\theta) \)[/tex], where ( F ) is the magnitude of the applied force and [tex]\( \theta \)[/tex] is the angle of inclination of the slope. Substituting the given values, [tex]\( F_{\text{parallel}} = 182 \sin(22.3^\circ) \)[/tex]. This force component is larger than the force of static friction, causing the block to overcome friction and slide downward.
In summary, when a force exceeding 182 N is applied parallel to the slope, the block moves downward along the slope due to gravity overcoming friction. The magnitude of the applied force, along with the angle of inclination of the slope, determines the motion of the block. Therefore, option c) The block moves downward along the slope accurately describes the resulting motion. The correct option is c.
Complete Question:
A 59.1 kg block rests on a sloped surface inclined at 22.3°. If a force exceeding 182 N is applied parallel to the slope, what will be the block's resulting motion?
a) The block remains at rest.
b) The block moves upward along the slope.
c) The block moves downward along the slope.
d) The block oscillates.
Answer:
In this case, the block will slide down the slope when a force exceeding 182 N is applied parallel to the slope.
The answer is option~c
Explanation:
To determine the movement of the block on the slope when a force exceeding 182 N is applied parallel to the slope, we need to consider the forces acting on the block and the friction force.
1. Identify the forces acting on the block:
- Force of gravity pulling the block downward (mg)
- Normal force perpendicular to the slope (N)
- Applied force parallel to the slope (Fapplied)
- Friction force opposing motion along the slope (f)
2. Resolve the forces into components parallel and perpendicular to the slope:
- Force parallel to the slope = Fapplied
- Force perpendicular to the slope = mg * cos(θ), where θ = 22.3°
3. Calculate the maximum static friction force that can prevent the block from sliding using the formula:
f(max) = μs * N, where μs is the coefficient of static friction
4. Determine the friction force required to keep the block stationary:
f = Fapplied + mg * sin(θ), where θ = 22.3°
5. Compare the friction force to the maximum static friction force:
- If f < f(max), the block remains stationary.
- If f > f(max), the block will slide in the direction of the applied force.
6. Given that the applied force exceeds 182 N, the block will slide down the slope (c) since the friction force required to keep it stationary is exceeded.
Therefore, the block will slide down the slope when a force exceeding 182 N is applied parallel to the slope.
The answer is option~c
