Question

A hockey stick exerts a horizontal force of magnitude 3.0 N on a 0.25-kg puck. While under the influence of the stick's force (and possibly some other forces), the puck moves 2.0 m. a) Is it possible that the stick does 6.0 J of work on the puck? If so, what is the angle between the stick's force and the puck's displacement while the force acts? b) Is it possible that the stick does - 3.0 J of work on the puck? If so, what is the angle between the stick's force and the puck's displacement? c) Is it possible that the stick does -6.0 J of work on the puck? If so, what is the angle between the stick's force and the puck's displacement? d) Is it possible that the stick does 6.0 J of work on the puck? If so, what is the angle between the stick's force and the puck's displacement? e) If the puck possesses an initial speed of 4.0 m/s, and then 4.0 J of work is done on the puck, find its final speed. Ignore friction. f) If the puck possesses an initial speed of 4.0 m/s, and then -2.0 J of work is done on the puck, find its final speed. Ignore friction.

Answer 1

work done = = F s cos

where is angle between direction of force and displacement.

(a) work done = F s cos = 3 2 cos = 6 J , i.e. cos = 1 hence = 0^o

angle between stick's force and puck's displacement is zero, i.e. force and displacements are parallel , same direction.

(b) work done = F s cos = 3 2 cos = - 3 J , i.e. cos = -1/2, hence = 120^o

angle between stick's force and puck's displacement is 120^o

(c) work done = F s cos = 3 2 cos = - 6 J , i.e. cos = -1 hence = 180^o

angle between stick's force and puck's displacement is 180^o .

( In this case puck is approaching the stick with high speed and puck pushes the stick backward,

while stick exerting some force on puck that will reduce the speed of puck but stick will move backward with puck stick to it. )

(d) work done 6 J is already done in case (a)

(e) By work-energy theorem,

(1/2) m v_f² = (1/2) m v_i² + (work done) = (1/2)0.2544 + 4 = 6 J

where v_i is initial speed and v_f is final speed of puck

hence v_f = (12/0.25)^1/2 = 6.93 m/s

(f) By work-energy theorem,

(1/2) m v_f² = (1/2) m v_i² + (work done) = (1/2)0.2544 - 2 = 0

hence v_f = 0 m/s