Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

A box is sliding down an incline tilted at an angle 14. 0° above horizontal. The box is sliding down the incline at a speed of 1. 70 m/s. The coefficient of kinetic friction between the box and the incline is 0. 380. How far does the box slide down the incline before coming to rest?.

Sagot :

LammettHash

The net forces on the box acting parallel and perpendicular to the incline, respectively, are

∑ F[para] = mg sin(14.0°) - F[friction] = ma

∑ F[perp] = F[normal] - mg cos(14.0°) = 0

where m = mass of the box and a = its acceleration.

The second equation tells us

F[normal] = mg cos(14.0°)

so that friction has a magnitude of

F[friction] = 0.380 F[normal] = 0.380 mg cos(14.0°)

Solve the first equation for a :

mg sin(14.0°) - 0.380 mg cos(14.0°) = ma

a = g sin(14.0°) - 0.380 g cos(14.0°)

a ≈ -1.24 m/s²

Assuming the box slides down the incline with constant acceleration, solve for ∆x, the distance it slides before coming to rest:

0² - (1.70 m/s)² = 2a ∆x

∆x = -(1.70 m/s)² / (2 (-1.24 m/s²))

∆x ≈ 1.16 m

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.