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Question

(D) For a block moving at a uniform velocity,the net force acting on it must be zero.Let $F$ be the applied force. The vertical forces are the normal

Vedclass · Accepted Answer

$\frac{\mu mgd}{\cos 	heta + \mu \sin 	heta}$

Vedclass · Answer

(D) For a block moving at a uniform velocity,the net force acting on it must be zero.Let $F$ be the applied force. The vertical forces are the normal force $N$,the vertical component of the applied force $F \sin 	heta$,and the weight $mg$. Thus,$N + F \sin 	heta = mg$,which gives $N = mg - F \sin 	heta$.The horizontal forces are the horizontal component of the applied force $F \cos 	heta$ and the kinetic friction $f_k = \mu N$. For uniform velocity,$F \cos 	heta = f_k = \mu(mg - F \sin 	heta)$.Solving for $F$: $F \cos 	heta = \mu mg - \mu F \sin 	heta \implies F(\cos 	heta + \mu \sin 	heta) = \mu mg \implies F = \frac{\mu mg}{\cos 	heta + \mu \sin 	heta}$.The work done $W$ during displacement $d$ is $W = (F \cos 	heta) 	imes d$. However,since the work done by the applied force is requested,$W = F \cdot d \cdot \cos 	heta$ is not the standard interpretation here; usually,it refers to the work done by the applied force $F$ in the direction of displacement. Given the options,the work done by the applied force is $W = F \cdot d = \frac{\mu mgd}{\cos 	heta + \mu \sin 	heta}$.

$A$ block of mass $m$ is pulled along a horizontal surface by applying a force at an angle $\theta$ with the horizontal. The friction coefficient between the block and the surface is $\mu$. If the block travels at a uniform velocity,then the work done during its displacement $d$ is:

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