A block of mass 25 kg is pulled along a horizontal surface by a force at an angle 45^circ$45^{circ}$ with the horizontal. The friction coefficient between the block and the surface is 0.25. The work done for a displacement of 5 m of the block with uniform velocity is:

### Related Formula f_s leq mu_s N$$f_s \leq \mu_s N$$ F_c = mromega^2$$F_c = mr\omega^2$$ ### Core Logic

To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion. f = momega^2 r$f = m\omega^2 r$ The normal force on the flat disc is N = mg$N = mg$. The maximum static friction is f_textmax = mu N = mu mg$f_{\text{max}} = \mu N = \mu mg$. For no slipping: m r omega^2 leq mu mg$$m r \omega^2 \leq \mu mg$$ omega^2 leq fracmu gr$$\omega^2 \leq \frac{\mu g}{r}$$ omega_textmax = sqrtfracmu gr$$\omega_{\text{max}} = \sqrt{\frac{\mu g}{r}}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws Of Motion