A block of mass 2 ,kg is pulled at a constant | vedclass

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(A) Given:Mass of the block,$m = 2 \,kg$Angle of inclination,$	heta = 30^{\circ}$Distance moved,$d = 4 \,m$Acceleration due to gravity,$g = 10 \,ms^{

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) Given:Mass of the block,$m = 2 \,kg$Angle of inclination,$	heta = 30^{\circ}$Distance moved,$d = 4 \,m$Acceleration due to gravity,$g = 10 \,ms^{-2}$Since the block is pulled at a constant speed,the net force acting on the block along the inclined plane is zero.Therefore,the tension $T$ in the rope must balance the component of the gravitational force acting down the plane:$T = mg \sin 	heta$$T = 2 	imes 10 	imes \sin 30^{\circ}$$T = 20 	imes 0.5 = 10 \,N$The work done by the tension $W$ is given by:$W = T 	imes d 	imes \cos(0^{\circ})$$W = 10 \,N 	imes 4 \,m 	imes 1$$W = 40 \,J$

$A$ block of mass $2 \,kg$ is pulled at a constant speed with a taut rope along a frictionless plane that is inclined at $30^{\circ}$. Then the work done by the tension in the rope in pulling it a distance $4 \,m$ along the inclined plane in joule is (acceleration due to gravity $= 10 \,ms^{-2}$)

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