Acceleration of a body sliding down a smooth incline A block slides down a frictionless plane inclined at angle θ to the horizontal. What is its acceleration component along the plane?

Difficulty: Easy

Correct Answer: g sin θ

Explanation:


Introduction / Context:
Resolving gravitational acceleration into components along and perpendicular to an inclined plane is a fundamental step in many statics and dynamics problems. With no friction, only the component of weight along the plane accelerates the block.


Given Data / Assumptions:

  • Inclination angle θ to the horizontal.
  • Frictionless (smooth) plane.
  • Gravity g acts vertically downward.


Concept / Approach:

Resolve weight W = m g into components: perpendicular to plane (m g cos θ) and along plane (m g sin θ). With no friction, the equation of motion along the plane is m a = m g sin θ, leading directly to a = g sin θ.


Step-by-Step Solution:

Choose axes: x along plane downward; y normal to plane.Along y: N − m g cos θ = 0 ⇒ N = m g cos θ.Along x: m a = m g sin θ ⇒ a = g sin θ.


Verification / Alternative check:

Limiting cases: θ = 0° ⇒ a = 0; θ = 90° ⇒ a = g. Both agree with intuition.


Why Other Options Are Wrong:

g is the full vertical acceleration, not along the plane; g cos θ is the normal component; g tan θ is not a component of gravitational acceleration; zero would require θ = 0° or friction balancing, which is not the case here.


Common Pitfalls:

Using g cos θ for the along-plane component; forgetting to define axes parallel and perpendicular to the plane.


Final Answer:

g sin θ

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