Question

In 1990 Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work did Arfeuille do on the object? (b) What magnitude force did he exert on the object during the lift, assuming the force was constant?

Answer 1

(a). The work done is 472 J.

(b). The force exerted is 2.76 kN.

Step-by-step explanation:

Given that,

Distance = 17.1 cm

Mass of object = 281.5 kg

(a). We need to calculate the work done

Using formula of work done

`W=F_(net)\dotc d`

`W=mg\cdot d`

Where, m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

`W=281.5*9.8*17.1*10^(-2)`

`W=472\ J`

The work done is 472 J.

(b). We need to calculate the force

Using action reaction principle

`F=F'`

`F=mg`

Put the value into the formula

`F=281.5*9.8`

`F=2.76\ kN`

The force exerted is 2.76 kN.

Hence, (a). The work done is 472 J.

(b). The force exerted is 2.76 kN.