A 500N box rests on a horizontal floor. A constant horizontal force is exerted on the box so that it moves through 8m. If the coefficient of kinetic  friction between the floor and the box is 0.22, calculate the workdone on the box
 

 

 

Thephysicsteacher

4 years ago

I think the answer is A.

Theoneontheceiling

1 year ago

Identify the given values. Force (F) = 500 N, distance (d) = 8 m, coefficient of kinetic friction (μ) = 0.22.

Calculate the frictional force (Ff). Ff = μ * Normal force. Since the box is on a horizontal surface, the normal force equals the weight of the box. We are not given the mass, so we cannot calculate the weight directly. However, the problem states a constant horizontal force is applied, implying the applied force overcomes friction. Therefore, the frictional force is 500N * 0.22 = 110 Ν.
(This is where I think the question has problems because it tells us to calculate the total workdone and it stops at 110N)

Here is the workdone though:


Calculate the net force (Fnet). Fnet = Applied force - Frictional force = 500 N-110 N = 390 Ν.

Calculate the work done (W). Work is the product of force and distance. W = Fnet * d = 390 N * 8 m = 3120 J.

JazrealMEMEs

3 years ago

Calculation please

a7046146828

3 months ago

Workdone here = coefficient of kinetic × Force × distance
Work = 0.22 × 500 × 8
Work = 880J