The number 3252.6 has 5 significant figures
Work-Energy :W = 1/2 m ( Vf^2 -Vo^2 )
Vo = 24.0 m/s Initial speed
Vf = 27.5 m/s Final speed
W = 1/2 m ( Vf^2 -Vo^2 )
160 kj = 1/ 2 m ( 27.5^2 -24.0 ^2)
160kj = 4680 x m
convert kilo joules to jeoules 160000 j = 4689 xm
m = 160000 j/4689
m = 34.18 kg
Remember what acceleration is? It's how fast the speed is changing over a period of time.

So in dt = 2 secs, she sped up from 6 to 10 m/s and so dv = (10-6) m/s.
Now that you know dv and dt, you can calculate the acceleration a!
Answer:
Answered
Explanation:
1 and 3 are necessary
Every bit of force applied to the bumper will be transmitted to the cart EXCEPT for the force needed to accelerate the bumper. This is the net force on the bumper.
If the bumper was heavy then a significant amount of force might be needed to accelerate the bumper so the amount transmitted to the cart would be substantially reduced.
If the net force on the bumper is small then the amount transmitted to the cart is almost the entire force applied.