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:
\( \sqrt{5} \, J \)
\( 5 \times 10^3 \, J \)
\( 10 \, J \)
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:
\[ \text{Work} = \mu_k \times N \times d = 0.1 \times 5 \times 10 = 5 \, J. \]
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