Final answer:
The calculated coefficient of friction μ is approximately 0.0408, which does not match the provided options. This determination requires using Newton's second law and the definition of frictional force. There might be an error in the options or an additional force acting on the crate.
Step-by-step explanation:
The question asks for the coefficient of friction (μ) between a crate and the surface. The crate has a mass of 50 kg and is acted upon by a constant horizontal force of 20 N, which brings it to a speed of 4 m/s after 5 seconds. To solve this, we can use Newton's second law of motion and the definition of the coefficient of friction.
Firstly, we'll calculate the acceleration (a) of the crate using the formula a = Δv/Δt, where Δv is the change in velocity and Δt is the change in time. Since the crate starts from rest and achieves a velocity of 4 m/s in 5 s, the acceleration is 4 m/s / 5 s = 0.8 m/s².
Next, we apply Newton's second law, F = ma, to find the net force. The mass (m) is 50 kg and the acceleration (a) is 0.8 m/s², so F_net = 50 kg * 0.8 m/s² = 40 N.
Since the applied force (F_app) is 20 N and is less than the net force, there must be a force of friction (F_friction) opposing the motion. We can calculate this frictional force as F_net - F_app = 40 N - 20 N = 20 N.
The force of friction can also be expressed as F_friction = μN, where N is the normal force. The normal force, in this case, is equal to the weight of the crate, which is mg = 50 kg * 9.8 m/s² = 490 N. Thus, μ = F_friction / N = 20 N / 490 N ≈ 0.0408.
However, none of the options given (a) μ = 0.10, (b) μ = 0.20, (c) μ = 0.30, (d) μ = 0.40 match our calculated value. There might be a mistake in the question or the options provided. If the question and options are correct, then it could imply an additional force acting on the system that hasn't been accounted for or a misinterpretation of the coefficients.