---
title: "A horizontal force of \\(110 \\, \\text{N}\\) is applied to a \\(12 \\, \\text{kg}\\) object, moving it \\(6 \\, \\text{m}\\) on a horizontal surface where the kinetic friction coefficient is \\(\\mu_k = 0.25\\). The object then slides up a \\(17^\\circ\\) inclined plane. Assuming the \\(110 \\, \\text{N}\\) force is no longer acting on the incline, and the coefficient of kinetic friction there is \\(\\mu_k = 0.45\\), calculate the distance the object will slide on the incline."
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url: "https://nerd-notes.com/ubq/23124/"
date_modified: "2025-12-14T02:40:56+00:00"
---

# A horizontal force of \(110 \, \text{N}\) is applied to a \(12 \, \text{kg}\) object, moving it \(6 \, \text{m}\) on a horizontal surface where the kinetic friction coefficient is \(\mu_k = 0.25\). The object then slides up a \(17^\circ\) inclined plane. Assuming the \(110 \, \text{N}\) force is no longer acting on the incline, and the coefficient of kinetic friction there is \(\mu_k = 0.45\), calculate the distance the object will slide on the incline.

A horizontal force of \(110 \, \text{N}\) is applied to a \(12 \, \text{kg}\) object, moving it \(6 \, \text{m}\) on a horizontal surface where the kinetic friction coefficient is \(\mu_k = 0.25\). The object then slides up a \(17^\circ\) inclined plane. Assuming the \(110 \, \text{N}\) force is no longer acting on the incline, and the coefficient of kinetic friction there is \(\mu_k = 0.45\), calculate the distance the object will slide on the incline.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/23124/*