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

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

$W = 2.20 \times 10^5 J$

also we know that

$d = 8.00 m$

now we will have

$2.20 \times 10^5 = F(8.00)$

$F = \frac{2.20 \times 10^5}{8}$

$F = 27500 N$

so it will have 27500 N force on it