The work done in pulling up a block of wood w | vedclass

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(B) The force required to pull the block up the smooth inclined plane at a constant velocity is equal to the component of the gravitational force acti

Vedclass · Accepted Answer

$5.17$

Vedclass · Answer

(B) The force required to pull the block up the smooth inclined plane at a constant velocity is equal to the component of the gravitational force acting down the plane,which is $F = mg \sin 	heta$.Given that the weight of the block $mg = 2 \, kN = 2000 \, N$,the length of the plane $s = 10 \, m$,and the angle of inclination $	heta = 15^\circ$.The work done $W$ is given by the product of the force and the displacement: $W = F 	imes s = (mg \sin 	heta) 	imes s$.Substituting the values: $W = 2000 	imes \sin(15^\circ) 	imes 10$.Since $\sin(15^\circ) \approx 0.2588$,we get $W = 2000 	imes 0.2588 	imes 10 = 5176 \, J$.Converting to kilojoules: $W = 5.176 \, kJ \approx 5.17 \, kJ$.

The work done in pulling up a block of wood weighing $2 \, kN$ for a length of $10 \, m$ on a smooth plane inclined at an angle of $15^\circ$ with the horizontal is ............. $kJ$.

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