Question

During tug-of-war, team A does 2.20*10^5 J of work in pulling team B 8.00m. What average force did team A exert

Answer 1

The average force exerted by team A during the tug-of-war is 27,500 N, calculated by dividing the work done (2.20×105 J) by the distance pulled (8.00 m).

Step-by-step explanation:

During a tug-of-war, team A does 2.20×105 J of work in pulling team B 8.00 m. To find the average force that team A exerted, we can use the work done on team B. Work is defined as the product of force and the distance over which the force acts, and it's expressed as Work = Force × Distance. In this situation, we can rearrange the formula to solve for force: Force = Work / Distance.

Given the work done, 2.20×105 J, and the distance of 8.00 m, the average force exerted by team A can be calculated as:

Force = 2.20×105 J / 8.00 m = 2.75×104 N

Therefore, the average force exerted by team A during the tug-of-war is 27,500 N.

Answer 2

As we know that work done is defined as product of force and displacement

so here we will have

`W = F.d`

so here we know that

`W = 2.20 * 10^5 J`

also we know that

`d = 8.00 m`

now we will have

`2.20 * 10^5 = F(8.00)`

`F = (2.20 * 10^5)/(8)`

`F = 27500 N`

so it will have 27500 N force on it