Question

A horizontal force is exerted on a 20 kg box to slide it up on an inclined plane with an angle of 30°. The frictional force retarding the motion is 80 N. If the box moves with a constant speed, then the magnitude of the force is:(Take g=10 ms^-2)

• A. 50\(\sqrt{2}\) N
• B. 100 N
• C. 80\(\sqrt{3}\) N
• D. 100\(\sqrt{2}\) N
• E. 120\(\sqrt{3}\) N

Answer 1

Answer: (E)

Given:

• Mass of the box, \( m = 20 \, \text{kg} \)
• Incline angle, \( \theta = 30^\circ \)
• Frictional force, \( f = 80 \, \text{N} \) (retarding motion)
• Box moves with constant speed (acceleration \( a = 0 \))
• Gravitational acceleration, \( g = 10 \, \text{ms}^{-2} \)

Step 1: Resolve Forces Along the Incline

The component of gravitational force along the incline is:

\[ mg \sin \theta = 20 \times 10 \times \sin 30^\circ = 200 \times 0.5 = 100 \, \text{N} \]

Step 2: Apply Equilibrium Condition

Since the box moves with constant speed, the net force along the incline is zero. Thus:

\[ F_{\text{applied}} \cos \theta = mg \sin \theta + f \]

Here, \( F_{\text{applied}} \) is the horizontal force we need to find. Resolving it along the incline:

\[ F_{\text{applied}} \cos 30^\circ = 100 + 80 \]

\[ F_{\text{applied}} \times \frac{\sqrt{3}}{2} = 180 \]

\[ F_{\text{applied}} = \frac{180 \times 2}{\sqrt{3}} = \frac{360}{\sqrt{3}} = 120\sqrt{3} \, \text{N} \]

Conclusion:

The magnitude of the horizontal force is \( 120\sqrt{3} \, \text{N} \).

Answer: \(\boxed{E}\)

Answer 2

1. Define variables and given information:

m = 20 kg (mass of the box)

θ = 30° (angle of incline)

f = 80 N (frictional force)

g = 10 m/s² (acceleration due to gravity)

F = ? (magnitude of the horizontal force)

Constant speed implies zero acceleration.

2. Resolve forces along the incline:

Component of horizontal force along the incline: \[F \cos θ = F \cos 30° = F (\sqrt{3}/2)\]

Component of weight along the incline: \[mg \sin θ = (20)(10) \sin 30° = 100\] N

Frictional force: \(f = 80\) N (opposing motion)

3. Apply Newton's second law along the incline:

Since the box moves with constant speed, the net force along the incline is zero:

\[F \cos θ - mg \sin θ - f = 0\]

\[F (\sqrt{3}/2) - 100 - 80 = 0\]

\[F (\sqrt{3}/2) = 180\]

\[F = \frac{180 \times 2}{\sqrt{3}} = \frac{360}{\sqrt{3}} = \frac{360 \sqrt{3}}{3} = 120\sqrt{3}\]