Question

A 925 n crate is being pulled across a level floor by a 325 n force exerted at an angle of 25 degrees above horizontal. If the coefficient of friction between the crate and the floor is 0.25, what is its acceleration?

Answer 1

The acceleration of the crate is 0.067 m/s².

Step-by-step explanation:

To find the acceleration of the crate, we first need to calculate the friction force acting on it. Since the crate is being pulled across a level floor, the normal force is equal to the weight of the crate, which is given by the formula W = mg. Then we can use the equation f = μN to find the friction force, where μ is the coefficient of friction and N is the normal force.

The normal force is equal to the weight, so N = 925 N. Substituting the values into the equation, we have f = 0.25 * 925 N = 231.25 N.

Next, we can calculate the net force acting on the crate by decomposing the applied force into its horizontal and vertical components. The horizontal component is calculate as Fx = F * cos(angle), and the net force is given by Fnet = Fx - f.

Substituting the values, we have Fx = 325 N * cos(25°) = 293.12 N, and Fnet = 293.12 N - 231.25 N = 61.87 N.

Finally, we can use Newton's second law, Fnet = ma, to find the acceleration. Rearranging the formula, we have a = Fnet / m. Substituting the values, we get a = 61.87 N / 925 kg = 0.067 m/s².