Question

A box of mass 17.6 kg with an initial velocity of 2.25 m/s slides down a plane, inclined at 19◦ with respect to the horizontal. The coefficient of kinetic friction is 0.48. The box stops after sliding a distance x. 17.6 kg µk = 0.48 2.25 m/s 19◦

How far does the box slide? The acceleration due to gravity is 9.8 m/s 2 . The positive x-direction is down the plane. Answer in units of m.

Answer 1

The sliding distance of the box can be calculated using the principles of conservation of energy and the work-energy theorem, considering the initial kinetic energy and the work done against the force of kinetic friction.

Step-by-step explanation:

Calculating the Sliding Distance of a Box on an Inclined Plane

To determine how far the box slides, we can use the principles of conservation of energy and the work-energy theorem. The initial kinetic energy of the box is transformed into work done against friction. The work done by friction is equal to the force of friction times the distance the box slides. We start with calculating the force of kinetic friction, which is μ_k (coefficient of kinetic friction) times the normal force. The normal force is the component of the box's weight perpendicular to the inclined plane, calculated as m*g*cos(θ), where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the incline.

With given values: m = 17.6 kg, μ_k = 0.48, θ = 19°, βi = 2.25 m/s, and g = 9.8 m/s², we can calculate the force of kinetic friction (ƒ_k).

The component of gravity along the incline is m*g*sin(θ), and we know that the box stops when its initial kinetic energy is equal to the work done by friction. So, from the equation:

Kinetic Energy_initial = Work_friction,

½*m*βi^2 = ƒ_k * distance,

½*17.6 kg*(2.25 m/s)^2 = (0.48*17.6 kg*9.8 m/s²*cos(19°)) * distance,

We can then solve for the distance the box slides. After calculation, we obtain the sliding distance x.

Answer 2

2.01 m

Step-by-step explanation:

Given parameters:

Mass, m = 17.6 kg

Velocity, v = 2.25 m/s

Degree = 19°

Coefficient of kinetic friction,
`\mu_k` = 0.48

To find the distance, we need to use Work-Energy principle.

So, total energy must be equal to the work done by friction. Here total energy is sum of the initial kinetic energy and the loss in potential energy.

`E_K = (1)/(2)mv^2\\E_P = mgxsin(\theta)\\W_(fr) = \mu_k mgcos(\theta)x\\`

So,
`(1)/(2)mv^2 + mgxsin(\theta) = \mu_k mgcos(\theta)x`

`x = ((1)/(2)mv^2)/(\mu_k mgcos(\theta)-mgsin(\theta)) = ((1)/(2)* 2.25^2)/(0.48*9.8*cos(19)-9.8*sin(19)) = 2.01` m