Question

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60° by a force of 10 N parallel to the inclined surface as shown in the figure. When the block is pushed up by 10 m along the inclined surface, the work done against frictional force is:

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

Answer 1

The Correct Option is C

The work done against the frictional force can be calculated using the formula:

\[ \text{Work} = \mu_k \times N \times d \]

Where:
- \( \mu_k = 0.1 \) (the coefficient of kinetic friction),
- \( N = mg \cos \theta \) (the normal force),
- \( d = 10 \, m \) (the distance moved along the inclined plane).

First, we find the normal force:

\[ N = mg \cos(60^\circ) = 1 \times 10 \times \frac{1}{2} = 5 \, N. \]

Now we can calculate the work done against friction:

\[ \text{Work} = \mu_k \times N \times d = 0.1 \times 5 \times 10 = 5 \, J. \]

Answer 2

The problem asks for the calculation of the work done against the frictional force when a block of mass 1 kg is pushed up an inclined surface. We are provided with the mass of the block, the angle of inclination, the applied force, the distance moved, and the coefficient of kinetic friction.

Concept Used:

The work done against a constant force is the product of the magnitude of the force and the distance moved along the direction of the force. The work done against the kinetic frictional force (\(W_{\text{friction}}\)) is given by:

\[ W_{\text{friction}} = f_k \cdot d \]

where \( d \) is the distance the object moves and \( f_k \) is the magnitude of the kinetic frictional force. The kinetic frictional force is defined as:

\[ f_k = \mu N \]

Here, \( \mu \) is the coefficient of kinetic friction and \( N \) is the normal force. For an object on an inclined plane, the normal force is the component of the gravitational force that is perpendicular to the plane's surface:

\[ N = mg \cos\theta \] where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of inclination.

Step-by-Step Solution:

Step 1: List all the given physical quantities from the problem statement.

Mass of the block, \( m = 1 \, \text{kg} \)
Angle of inclination, \( \theta = 60^\circ \)
Coefficient of kinetic friction, \( \mu = 0.1 \)
Distance pushed along the incline, \( d = 10 \, \text{m} \)
We will use the standard approximation for acceleration due to gravity, \( g \approx 10 \, \text{m/s}^2 \).

Step 2: Calculate the normal force (\(N\)) acting on the block.

The normal force is perpendicular to the inclined surface and balances the perpendicular component of the gravitational force. Its magnitude is calculated as:

\[ N = mg \cos\theta \]

Substituting the given values:

\[ N = (1 \, \text{kg}) \times (10 \, \text{m/s}^2) \times \cos(60^\circ) \]

Since \( \cos(60^\circ) = 0.5 \):

\[ N = 10 \times 0.5 = 5 \, \text{N} \]

Step 3: Calculate the magnitude of the kinetic frictional force (\(f_k\)).

The kinetic frictional force is given by the product of the coefficient of kinetic friction and the normal force.

\[ f_k = \mu N \]

Substituting the values of \( \mu \) and \( N \):

\[ f_k = 0.1 \times 5 \, \text{N} = 0.5 \, \text{N} \]

Step 4: Calculate the work done against the frictional force.

The work done against friction is the frictional force multiplied by the distance over which it acts. The frictional force opposes the motion of the block up the incline.

\[ W_{\text{against friction}} = f_k \times d \]

Final Computation & Result:

Substituting the values of the frictional force and the distance:

\[ W_{\text{against friction}} = (0.5 \, \text{N}) \times (10 \, \text{m}) \] \[ W_{\text{against friction}} = 5 \, \text{J} \]

The work done against the frictional force is 5 J.