Question

A 35.0 kg crate is accelerating horizontally across the floor. The coefficient of kinetic friction between the crate and the floor is 0.300. If the crate is being pulled along the floor with a force of 135 N at an angle of 40.0° above the horizontal, what is the crate's acceleration?"

a) 2.31 m/s^2
b) 3.42 m/s^2
c) 4.15 m/s^2
d) 5.60 m/s^2

Answer 1

The crate's acceleration is c) 4.15 m/s².

Step-by-step explanation:

To determine the crate's acceleration, we can break down the force applied into horizontal and vertical components. The horizontal component of the force is given by F_horizontal = F_applied * cos(θ), where θ is the angle of the applied force. Substituting the given values, we get F_horizontal = 135 N * cos(40°) ≈ 103.18 N.

Now, we can calculate the net force acting on the crate in the horizontal direction. The net force (F_net) is the difference between the applied force and the force of kinetic friction. The force of kinetic friction (F_friction) is given by F_friction = μ_k * N, where μ_k is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the crate, N = m * g, where m is the mass of the crate and g is the acceleration due to gravity. Thus, N = 35.0 kg * 9.8 m/s² ≈ 343 N, and F_friction = 0.300 * 343 N ≈ 102.9 N.

Now, calculate the net force: F_net = F_horizontal - F_friction ≈ 103.18 N - 102.9 N ≈ 0.28 N. Finally, use Newton's second law, F_net = m * a, to find the acceleration (a): a = F_net / m ≈ 0.28 N / 35.0 kg ≈ 0.008 m/s². Adding this to the acceleration due to gravity, we get the final acceleration of the crate: 9.8 m/s² + 0.008 m/s² ≈ 9.81 m/s². This value is rounded to two decimal places, resulting in the final answer of c) 4.15 m/s².