[xxxtazxxx]

A person pushes a 19.8-kg shopping cart at a constant velocity for a distance of 34.2 m on a flat horizontal surface. She pushes in a direction 24.4 degrees below the horizontal. A 50.4-N frictional force opposes the motion of the cart.
(A) : What is the magnitude of the force that the shopper exerts?
(B) : Determine the work done by - the pushing force
- the frictional force
- the gravitational force

[terryg752]

Since there is no acceleration, net force = 0

Let F be the force applied.

Component of this force along the surface = F cos 24.4 = 50.4

From this equation, find the value of F

Work done by pushing force = Force times distance = 50.4 times 34.2

Work done by Friction = same as above.

[nurat]

Magnitude of force used is frictional force.
THE WORK DONE-pushing force in frictional is force times distance 19.8 times 34.2=6771.6 J TIMES 10=67716

[Unknown008]

Magnitude of force used is frictional force.
THE WORK DONE-pushing force in frictional is force times distance 19.8 times 34.2=6771.6 J TIMES 10=67716

That's not completely right.

If you looked at terryg752's answer, you'll see that the magnitude of the applied force is not equal to frictional force. The horizontal component of the applied force is equal to the frictional force, not the applied force.

Secondly, the force is not 19.8 and you cannot say:



You did an arithmetic mistake and you inserted the 10 afterwards, which is not possible.