Power Needed for 200kg Glider to Reach 25m/s in 75m

[physics.mk]

A glider is launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. What average power must the winch supply in order to accelerate a 200 kg glider from rest to 25 m/s over a horizontal distance of 75 m. Air resistance and friction are negligible and tension in the winch is constant.

Homework Statement

v1=0m/s
v2=25 m/s
d=75 m
m=200 kg

Homework Equations

Power=change in work/change in time (p=w/t)
change in work=change in kinetic energy (w=k2-k1)
v=d/t

The Attempt at a Solution

w=k2-k1 (where k1=0 because initial velocity is 0)
w=k2 (0.5)(200)(25)^5 =6.25e4 J

now to get time: v=d/t ... t=d/v=75/25= 3s
sub w and t into the equation for power
p=6.25e4/3=20.8e3 W

but this isn't the right answer, the actual answer is 1.0e4 W .. i don't see where i went wrong.. help please

 

[Dick]

A glider is launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. What average power must the winch supply in order to accelerate a 200 kg glider from rest to 25 m/s over a horizontal distance of 75 m. Air resistance and friction are negligible and tension in the winch is constant.

Homework Statement

v1=0m/s
v2=25 m/s
d=75 m
m=200 kg

Homework Equations

Power=change in work/change in time (p=w/t)
change in work=change in kinetic energy (w=k2-k1)
v=d/t

The Attempt at a Solution

w=k2-k1 (where k1=0 because initial velocity is 0)
w=k2 (0.5)(200)(25)^5 =6.25e4 J

now to get time: v=d/t ... t=d/v=75/25= 3s
sub w and t into the equation for power
p=6.25e4/3=20.8e3 W

but this isn't the right answer, the actual answer is 1.0e4 W .. i don't see where i went wrong.. help please

Your determination of the time is wrong. You solved the problem as if the speed were a constant 25m/s. You know that's not true, right?

 

[physics.mk]

ah! now that you say it, thanks a lot