Question

a force of 57.1 newtons is applied for 0.977 meters, what work was done on the object? (1 point)
56.1 j
55.8 j
58.4 j
58.1 j

Explanation:

Step1: Recall the work formula

Work \(W = F\times d\), where \(F\) is the force and \(d\) is the distance.

Step2: Substitute the given values

Given \(F = 57.1\space N\) and \(d=0.977\space m\). Then \(W=57.1\times0.977\).

Step3: Calculate the product

\(57.1\times0.977=(57 + 0.1)\times0.977=57\times0.977+0.1\times0.977\).
\(57\times0.977=(60 - 3)\times0.977 = 60\times0.977-3\times0.977=58.62-2.931 = 55.689\).
\(0.1\times0.977 = 0.0977\).
\(W=55.689+0.0977=55.7867\approx55.8\space J\).

Answer:

55.8 J