A99.5 n grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 n. the acceleration of gravity is 9.81 m/s 2 . if all frictional forces are neglected and the cart starts from rest, what is the grocery cart’s final speed? answer in units of m/s. 003 (part 1 of 4) 10.0 points in the 1950’s, an experimental train that had a mass of 36300 kg was powered across a level track by a jet engine that produced a thrust of 4.28 × 105 n for a distance of 586 m. find the work done on the train. answer in units of j. 004 (part 2 of 4) 10.0 points find the change in kinetic energy. answer in units of j. 005 (part 3 of 4) 10.0 points find the final kinetic energy of the train if it started from rest. answer in units of j. 006 (part 4 of 4) 10.0 points find the final speed of the train assuming no friction. answer in units of m/s.

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

Since the cart was initially at rest, , so

(1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, :

So solving eq.(1) for v, we find the final speed of the cart:

2)

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

So the work done is

3)  

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

where

W is the work done

is the change in kinetic energy

Therefore, the change in kinetic energy is

4) 37.2 m/s

According to the work-energy theorem,

where

is the final kinetic energy of the train

is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

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