You apply a 56.3N force to a 4.2Kg cart towards the right. The cart accelerates towards the right at a rate of 0.8m/s2. What is the magnitude, in Newtons, of the net frictional force (the sum of the only other forces, besides the 56.3N one you apply)? Please enter only a numerical answer, no units necessary.

52,94N

Explanation:

Frictional force is the opposing force that is created between two surfaces that try to move in the same direction or that try to move in opposite directions. The main purpose of a frictional force is to create resistance to the motion of one surface over the other surface.

Calculation:

Le the value of frictional force be Fk, therefore acceleration a = (F- Fk)/m

where :

F is the applied force

m the mass of the cart

a = (F- Fk)/m = 0.8

F - Fk = 0.8m

Fk = F - 0.8m

Fk = 56.3 - (0.8×4.2)

Fk = 52.94 N

The magnitude, in Newtons, of the net frictional force is therefore 52.94N

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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nah.   i'm just sleepy.