Aluggage handler pulls a 20.0 kg suitcase up a ramp inclined at 33.0 ∘ above the horizontal by a force f⃗ of magnitude 161 n that acts parallel to the ramp. the coefficient of kinetic friction between the ramp and the incline is μk = 0.300. the suitcase travels 3.90 m along the ramp.

Complete question  

1) Calculate the work done on the suitcase by the force F

2) Calculate the work done on the suitcase by the gravitational force.  

3) Calculate the work done on the suitcase by the normal force.  

4) Calculate the work done on the suitcase by the friction force.  

5) Calculate the total work done on the suitcase.

The work done on the suitcase by the force F is 627.9Joules

The work done on the suitcase by the gravitational force is 416.32Joules  

The work done on the suitcase by the normal force is 0Joules  

The work done on the suitcase by the friction force is 192.32 Joules  

The total work done on the suitcase is 19.26Joules

The formula for calculating the work done by a body is expressed as:

Workdone = Force × distance

1) Work done on the suitcase = Horizontal force × distance traveled by the suitcase along the ramp.

Work done on the suitcase  = 161 × 3.90

Work done on the suitcase  = 627.9Joules

2) work done my gravitation= gravitational force × distance traveled along the ramp

Since gravitational force = mg

work done by gravitation = mghsinθ

Substitute the given parameters

work done by gravitation = 20×9.8×3.90sin33

work done by gravitation = 196×2.124

Work done by gravitation = 416.32Joules

3) If the force is normal to the plane then θ = 90degrees

Work done by normal force = 20×9.8×3.90sin0

work done by normal force = 196×0

Work done by normal force = 0Joules

4)  Frictional force = μRcosθ

Frictional force = μmgcosθ

Work done by friction = Frictional force × height

work done by friction = μmgcosθ × h

Work done by friction = 0.3 × 20 × 9.8 × cos33 × 3.9

Work done by friction = 229.32cos33

Work done by friction = 192.32 Joules

5) The total work done is the sum of all the forces acting on the suitcase.

Since the work done by gravitation and friction are both acting in the negative direction, hence;

Total work done =  627.9 + (-416.32) + 0 + (-192.32)

Total work done =  627.9 - 416.32 - 192.32

Total workdone on the suitcase = 19.26Joules

Learn more here: link

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .