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 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\( ^2 \), \( \mu_s = 0.1 \)]
Show Hint
- \( 5\sqrt{3} \) J
- \( 5 \) J
- \( 5 \times 10^3 \) J
- \( 10 \) J
The Correct Option is B
Solution and Explanation
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.
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