Question

A 17.4 kg block is dragged over a rough, horizontal surface by a constant force of 72.2 n acting at an angle of 28.1 ◦ above the horizontal. the block is displaced 16.9 m, and the coefficient of kinetic friction is 0.147. 17.4 kg µ = 0.147 72.2 n 28.1 ◦ find the work done by the 72.2 n force. the acceleration of gravity is 9.8 m/s 2 . answer in units of j.

Answer 1

The work done by the 72.2 N force is 75.73 J.

Step-by-step explanation:

The work done by a force can be calculated using the formula:

Work = Force * Distance * cos(θ)

Given: Force = 72.2 N, Distance = 16.9 m, θ = 28.1°

First, we need to find the horizontal component of the force:

F_horizontal = Force * cos(θ) = 72.2 N * cos(28.1°)

Next, we can calculate the work done by the force:

Work = F_horizontal * Distance = (72.2 N * cos(28.1°)) * 16.9 m

Substituting the values and calculating, we get:

Work = 75.73 J