what is the final speed ?

Question

a 95.1N grocery cart is pushed 17.9m along an aisle by a shopper who exerts a constant horizontal force of 31.2N. the acceleration of gravity is 9.8m/s^2. if all friction forces are neglected and the cart starts from rest, what is the final speed ?

Answer 1

the cart starts from rest, that means its velocity changes and there is an acceleration.

a = Fnet/m

only Fapply act on the cart so its the net force

a = Fapply/m
m = 95.1/9.8 = 9.7kg

a = 31.2N / 9.7kg
a = 3.22m/s^2

now that we have acceleration, find velocity

Vf^2 = 2ad + Vi^2
Vf^2 = 2(3.22)(17.9) + 0^2
Vf=10.73m/s

Answer 2

Whenever we fly, we should be concerned about flying the appropriate airspeed for the event involved. For example, many pilots fly a final approach airspeed they’ve become accustomed to regardless of its impact on where the airplane will touch down on the runway. They have become fond of a particular airspeed, regardless of its suitability. I’d like to challenge you to use a final approach airspeed closer to that which the airplane was certificated. I’m going to use the term Vref in place of “final approach speed.”

General aviation airplanes are certificated to Vref airspeeds 30% above stall speed in the landing configuration. With the landing gear down and flaps deployed, this is known as 1.3 times Vso. Unless there are performance charts for specific airplane weights, POH Vref airspeeds are normally based on Maximum Gross Weight (MGW). For example, a Cessna-182 POH indicates Vref, flaps down, of 70-80 knots at MGW. This is quite a speed range to enter the flare. However, the C-182 POH states that Vso is 49 knots. Therefore, at 1.3Vso, Vref can be 64 knots, a far cry from a span of 70-80 knots. Could the difference be a “lawyer thing”?
There is a rule of thumb that says that for each knot above Vref over the runway end numbers, the touchdown point will be an additional 100 feet down the runway than stated in the POH. Assuming the rule of thumb is correct and you are flying at POH Vref of 70-80 knots, instead of 64 knots, then you are landing 600 – 1600 fee

Answer 3

Hint: You don't need the acceleration of gravity for this.