Question

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of \( 60^\circ \) by a force of 10 N parallel to the inclined surface. When the block is pushed up by 10 m along the inclined surface, the work done against frictional force is: 
 

[Given: \( g = 10 \) m/s\( ^2 \), \( \mu_s = 0.1 \)] 
 

• A. \( 5\sqrt{3} \) J
• B. \( 5 \) J
• C. \( 5 \times 10^3 \) J
• D. \( 10 \) J

Hint:

The work done against friction is calculated as \( W = F_f d \), where \( F_f = \mu mg \cos \theta \).

Answer 1

The Correct Option is B

To determine the work done against frictional force, let's first analyze the forces acting on the block:

• The gravitational force component parallel to the incline: \( F_{\text{gravity}} = mg \sin \theta \), where \( m = 1 \) kg, \( g = 10 \) m/s\(^2\), and \( \theta = 60^\circ \). Thus, \( F_{\text{gravity}} = 1 \times 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \) N.
• The normal force \( N \) is equal to the gravitational force component perpendicular to the incline: \( N = mg \cos \theta = 1 \times 10 \times \frac{1}{2} = 5 \) N.
• The frictional force \( F_f \) is given by \( \mu \times N \), where \( \mu = 0.1 \). Thus, \( F_f = 0.1 \times 5 = 0.5 \) N.

Now, the work done against the frictional force when the block is pushed up the incline by 10 m is calculated as follows:

\[ \text{Work} = F_f \times \text{distance} = 0.5 \times 10 = 5 \text{ J} \]

Thus, the work done against the frictional force is \( 5 \) J.

Answer 2

Step 1: {Calculate normal force}
The normal force \( N \) on the inclined plane is: \[ N = mg \cos 60^\circ \] Step 2: {Determine frictional force}
\[ F_f = \mu N = \mu mg \cos 60^\circ \] Substituting values: \[ F_f = (0.1) (1) (10) \left( \frac{1}{2} \right) = 0.5 { N} \] Step 3: {Calculate work done against friction}
\[ W = F_f \times d \] \[ = 0.5 \times 10 = 5 { J} \] Thus, the correct answer is (B) 5 J.