Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

The coefficient of kinetic friction is 0.21, for a 8 kg block on the floor. It is being pulled by a 17 N force, what is the acceleration of the box?

Sagot :

The acceleration of the box is 0.025 m/s²

We'll begin by calculating the the frictional force

Coefficient of kinetic friction (μ) = 0.21

Mass (m) = 8 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 8 × 10 = 80 N

Frictional force (Fբ) =?

Fբ = μN

Fբ = 0.21 × 80

Fբ = 16.8 N

  • Next, we shall determine the net force acting on the box

Frictional force (Fբ) = 16.8 N

Force (F) = 17 N

Net force (Fₙ) =?

Fₙ = F – Fբ

Fₙ = 17 – 16.8

Fₙ = 0.2 N

  • Finally, we shall determine the acceleration of the box

Mass (m) = 8 Kg

Net force (Fₙ) = 0.2 N

Acceleration (a) =?

a = Fₙ / m

a = 0.2 / 8

a = 0.025 m/s²

Therefore, the acceleration of the box is 0.025 m/s²

Learn more: https://brainly.com/question/364384

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.