Question

A box of mass 3.0 kg

slides down a rough vertical wall. The gravitational force on the box is 29.4 N. When the box reaches a speed of 2.5 m/s, you start pushing on one edge of the box at a 45∘
angle (use degrees in your calculations throughout this problem) with a constant force of magnitude Fp= 23.0 N, as shown in (Figure 1). There is now a frictional force between the box and the wall of magnitude 13.0 N. How fast is the box sliding 2.8 s
after you started pushing on it?

Answer 1

To find the speed of the box 2.8 s after pushing it, we need to determine the net force acting on the box and use Newton's second law of motion.

Step-by-step explanation:

To find the speed of the box 2.8 s after pushing it, we need to determine the net force acting on the box and use Newton's second law of motion.

First, let's consider the forces acting on the box. We have the gravitational force (29.4 N) and the frictional force (13.0 N) acting in the opposite direction of motion. We also have the pushing force (23.0 N) acting at a 45° angle.

To find the net force, we need to resolve the pushing force into its horizontal and vertical components. The horizontal component of the pushing force is given by Fp*cos(45°) = Fp*cos(45°) = 16.3 N, and the vertical component is given by Fp*sin(45°) = Fp*sin(45°) = 16.3 N.

Now, we can calculate the net force in the horizontal direction. The net force in the horizontal direction is the difference between the horizontal component of the pushing force and the frictional force: Fnet = 16.3 N - 13.0 N = 3.3 N.

Finally, we can use Newton's second law of motion to find the acceleration of the box: a = Fnet/m = 3.3 N/3.0 kg = 1.1 m/s².

Now, we can use the kinematic equation to find the final speed of the box 2.8 s after pushing it:

v = u + at = 2.5 m/s + 1.1 m/s² * 2.8 s = 2.5 m/s + 3.08 m/s = 5.58 m/s.

Therefore, the box is sliding at a speed of 5.58 m/s 2.8 s after you started pushing on it.