GRE Physics Companion: Inclined-Plane Dynamics

Inclined-plane questions are usually component problems disguised as force problems. The fastest reliable strategy is to rotate the axes so that one axis lies along the plane, resolve the forces once, and then use Newton’s second law without changing coordinate systems mid-solution.

1 The component pair to know

For a plane inclined by angle 𝜃,

(1)

and

(2)

These are components of the Weight, not separate forces.

2 High-value results

For a frictionless block sliding down the plane,

(3)

For a simple block in static equilibrium with no other forces,

(4)

At impending slip,

(5)

For downhill kinetic sliding,

(6)

For uphill kinetic sliding, the acceleration points downhill with magnitude

(7)

3 Sign discipline

Pick a positive direction before writing equations. A negative result simply means the acceleration is opposite the assumed direction.

4 Worked GRE example 1: frictionless incline

A block slides from rest on a frictionless 37.0^∘ incline. Find the acceleration magnitude.

Immediately,

(8)

Therefore

(9)

The mass is irrelevant.

5 Worked GRE example 2: hanging mass versus incline block

A 4.00 kg block lies on a frictionless 30.0^∘ incline and is connected over an ideal Pulley to a hanging 3.00 kg mass. Which way does the system accelerate, and what is the acceleration magnitude?

Compare the two driving terms:

(10)

while

(11)

The hanging side wins. Thus

(12)

So

(13)

6 GRE speed questions

1. A block slides down a frictionless incline. If the incline angle increases, the acceleration magnitude (A) decreases (B) increases (C) remains constant (D) becomes zero.
2. A block rests on a simple rough incline at the threshold of downhill slipping. The coefficient of static friction is (A) sin 𝜃 (B) cos 𝜃 (C) tan 𝜃 (D) cot 𝜃.
3. A block slides downhill on an incline with kinetic friction. Which expression gives its acceleration magnitude? (A) g(sin 𝜃 + μ_k cos 𝜃) (B) g(sin 𝜃 − μ_k cos 𝜃) (C) g cos 𝜃 (D) μ_kg.
4. A horizontal force pushes a block toward the uphill direction on an incline that rises to the right. The force’s Normal component tends to (A) increase N (B) decrease N (C) leave N unchanged (D) eliminate gravity.
5. A block is moving uphill on a rough incline after being launched. Kinetic friction points (A) uphill (B) downhill (C) normal to the plane (D) vertically downward.
6. A block on an incline is connected to a hanging mass. Your equation gives a < 0 after you assumed the hanging mass moves downward. The correct interpretation is (A) the algebra is invalid (B) the system is in equilibrium (C) the acceleration is opposite the assumed direction (D) the tension is negative.

7 Answers and rationales

1. B. a = g sin 𝜃, and sin 𝜃 increases from 0 to 1 as the incline steepens from 0^∘ to 90^∘.
2. C. At impending slip, mg sin 𝜃 = μ_smg cos 𝜃.
3. B. Downhill gravity is opposed by uphill kinetic friction.
4. A. The horizontal push has a component into the plane, so the normal force increases.
5. B. Kinetic friction opposes the relative sliding, so it points downhill while the block slides uphill.
6. C. A negative result reverses the assumed acceleration direction; it is not an algebra failure.

References