A block (m=5.4 kg) is pulled by a force Fbar =18 Ni^ +7 N j^, where the x-axis i

Question

A block (m=5.4 kg) is pulled by a force Fbar =18 Ni^ +7 N j^, where the x-axis is the horizontal and the y-axis is the vertical on the figure. The block is pulled toward the positive x-direction a distance l = 11.4 mi^. How work is done? If the block was pulled from rest by this force over this distance, w is its speed after it is pulled the 11.4 m? If this is now repeated over a surface which has a kinetic coefficient of friction, muk = 0.26, w is the work done by friction over the same path? W is the speed of the block after the 11.4 m in this case?

Explanation / Answer

we have F=18i +7j (a) work is done W=F*11.4i=18*11.4=205.2J   (i*j=0) (b)initial velocity=0 conservation of energy final : W=m*v2/2 => v=8.72 m/s (c)the work done by the friction : normal force N=mg-7=45.92 N => frction force Ff=N*0.26=11.9 N => W'=Ff*11.4=136.1J (d)the speed is v' conservation of energy W-W'=m*v'2/2=> v'=5.06 m/s (d)the speed is v' conservation of energy W-W'=m*v'2/2=> v'=5.06 m/s

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