You push a 28kg wheelbarrow forward at a constant velocity with 600 n of force. how much is friction pushing backwards?

F = applied force on the wheelbarrow in forward direction = 600 N

f = frictional force pushing in backward direction of wheelbarrow = ?

m = mass of the wheelbarrow = 28 kg

Since wheelbarrow moves forward at a constant velocity

a = acceleration of the wheelbarrow in forward direction = 0 m/s²

force equation for the motion of wheelbarrow in forward direction is given as

F - f = ma

inserting the above values in the formula

600 - f = (28) (0)

600 - f = 0

f = 600 N

hence the frictional force comes out to be 600 N

higher potential

explanation:

because it is in a high speed so it must move into a of high potential

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle