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(D) The block is moving on a rough horizontal surface. The forces acting on the block are the normal force $N$, gravitational force $mg$, the pulling

Vedclass · Accepted Answer

$\frac{P}{\mu m g}$

Vedclass · Answer

(D) The block is moving on a rough horizontal surface. The forces acting on the block are the normal force $N$, gravitational force $mg$, the pulling force $F$, and the kinetic frictional force $fr_k$.For vertical equilibrium, the normal force is $N = mg$.The kinetic frictional force is given by $fr_k = \mu N = \mu mg$.Power $P$ is defined as the product of the applied force $F$ and the velocity $v$ in the direction of motion: $P = F v$.For the block to move at a constant velocity (which is its maximum velocity), the net force on the block must be zero.Therefore, the pulling force $F$ must be equal to the kinetic frictional force $fr_k$:$F = fr_k = \mu mg$.Substituting this into the power equation:$P = F v_{\max} = (\mu mg) v_{\max}$.Solving for $v_{\max}$:$v_{\max} = \frac{P}{\mu mg}$.

$A$ block of mass $m$ placed on a rough horizontal plane is pulled by a constant power $P$. The coefficient of friction between the block and the surface is $\mu$. The maximum velocity of the block will be

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