1.A golfer hits a shot to a green that is elevated 2.80 m above the point where

Question

1.A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 19.9 m/s at an angle of 32.0 above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

2.A 53.8-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.720 and 0.379, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?

PLZ help me with these two Questions. Thanks a lot.

Explanation / Answer

1 initial speed=19.9 m/s
angle=32 degrees
(here the question is asking for only speed. so we can solve it through energy conservation principle)

lets assume that the point where the ball is hit, is zero potnetial energy point.
let mass of the ball be m kg.
then total initial energy=initial kinetic energy=0.5*mass*speed^2=198*m

let the final speed be v.
at that moment it is 2.8 m above the base.

so total energy=0.5*mass*v^2+mass*g*2.8

=0.5*m*v^2+m*27.44

as there is no air resistance, using energy conservation principle:

198*m=27.44*m+0.5*m*v^2

v=18.469 m/s

hence it was travelling at 18.469 m/s.

2.

to just start the crate moving, the force has to be equal to maximum static friction force.
then force required=0.72*normal force=0.72*weight of the crate=0.72*53.8*9.8=379.61 N

now, once you get the crate moving, to move it across the deck at constant speed will mean that total force on the crate=0

hence force applied=kinetic friction on the crate

=> force applied=0.379*normal force=0.379*weight of the crate=0.379*53.8*9.8=199.82 N

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