The time it takes to move the box 7.18 m starting from rest is 3.545 s.
To determine the time it takes to move the box, we need to calculate the net force acting on the box and use Newton's second law of motion. First, let's find the force of friction.
The frictional force can be calculated using the formula: frictional force = coefficient of friction * normal force.
In this case, the normal force is equal to the weight of the box, which is given as 337 N.
Substituting the values, we get: frictional force = 0.583 * 337 N = 196.671 N.
The net force can be calculated by subtracting the frictional force from the horizontal force.
Net force = horizontal force - frictional force = 582 N - 196.671 N = 385.329 N.
Using Newton's second law of motion, we can calculate the acceleration of the box using the formula: net force = mass * acceleration.
Rearranging the formula to solve for acceleration, we get: acceleration = net force / mass = 385.329 N / 337 N = 1.143 m/s².
Finally, we can use the formula of motion, s = ut + 0.5at², to find the time it takes to move the box 7.18 m starting from rest.
Since the box starts from rest, the initial velocity (u) is 0. Substituting the values into the formula, we get: 7.18 = 0 + (0.5 * 1.143) * t².
Simplifying the equation, we have: 7.18 = 0.5715 * t². Dividing both sides by 0.5715, we get: t² = 12.56.
Taking the square root of both sides, we get: t = 3.545 s.
The probable question may be:
A box of books weighing 337 N is shoved across the floor of an apartment by a force of 582 N exerted downward at an angle of 25.3° below the horizontal.
The acceleration of gravity is 9.8 N.
If the coefficient of kinetic friction between box and floor is 0.583, how long does it take to move the box 7.18 m, starting from rest? Answer in units of s.