A box of mass m=2kg is pulled across a flat frictionless surface through a dista

Question

A box of mass m=2kg is pulled across a flat frictionless surface through a distance of d=1m with a force that makes an angle of 45 degrees above the horizontal with a magnitude of F=20N, see figure below. If the box's initial speed is v=0 m/s use the work-energy theorem, where v is the speed of the box and W total is the sum of all the work done by all the forces acting on the box, to find the box's final speed vf.

12. (10 points) A box of mass m 2 kg is pulled across a fat frictionless surface through a distance of d 1 m with a force that makes an angle of 0 45o above the horizontal with a magnitude of F 20 N, see figure below. If the box's initial speed is vi 0m/s use the work-energy theorem, total where v is the speed of the box and Wtot is the sum of all the work done by all the forces acting on the box, to find the box's final speed vf 2 kg

Explanation / Answer

Here,

F = 20 N at theta = 45 deg

Work done by displacement is only is Horizontal direction.

Fx = Fcos45

Fx = 20*cos45 = 14.14 N

Work done W = F.d = 14.14*1

W = 14.14 J

From work energy theorem,

W = (1/2)m*(vf^2 - vi^2)

14.14 = (1/2)*2*(vf^2 - 0)

vf = 3.76 m/s

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