Question

An object requires 500 N force to be pulled up on a \(30^\circ\) frictionless smooth inclined plane at a constant speed. Determine the weight of the object.

• A. \(500\sqrt{2} \, \text{N}\)
• B. \(1000 \, \text{N}\)
• C. \(1000\sqrt{2} \, \text{N}\)
• D. \(500\sqrt{3} \, \text{N}\)

Hint:

For objects moving at constant speed on a frictionless incline, applied force equals the component of weight along the incline: \(F = W \sin \theta\).

Answer 1

The Correct Option is B

Step 1: Think in terms of energy instead of forces.
Since the incline is frictionless and the object is pulled at constant speed, every joule of work the pulling force supplies goes straight into raising the object's height, nothing is lost anywhere else.

Step 2: Match the work done to the gain in potential energy.
If the object is pulled a length $L$ along the incline of angle $\theta = 30^\circ$, it rises a height $h = L\sin\theta$. Equating the work done by the pull to the gain in gravitational potential energy:
\[ F \cdot L = W \cdot h = W \cdot L\sin\theta \]

Step 3: Cancel the common length and solve for the weight.
The distance $L$ drops out of both sides, leaving
\[ F = W\sin\theta \implies W = \frac{F}{\sin\theta} = \frac{500}{\sin 30^\circ} = \frac{500}{0.5} \]

Step 4: Compute.
\[ W = 1000\ \text{N} \]

Step 5: Conclusion.
\[ \boxed{1000\ \text{N}} \]