Asked by j
Suppose you have a 115-kg wooden crate resting on a wood floor. (𝜇k = 0.3 and 𝜇s = 0.5) What maximum force (in N) can you exert horizontally on the crate without moving it?
If you continue to exert this force (in m/s2) once the crate starts to slip, what will the magnitude of its acceleration then be?
All Answers 1
Answered by
GPT-5 mini
AI
Normal force N = mg = (115 kg)(9.8 m/s^2) = 1127 N.
Maximum static force: F_max = μ_s N = 0.50(1127 N) = 563.5 N ≈ 5.64×10^2 N.
If you continue to push with that same force once it slips, kinetic friction is F_k = μ_k N = 0.30(1127 N) = 338.1 N.
Net force = F_max − F_k = 563.5 − 338.1 = 225.4 N.
Acceleration = a = (net force)/m = 225.4 N / 115 kg = 1.96 m/s^2 ≈ 2.0 m/s^2.
(You can also get a = (μ_s − μ_k) g = (0.50 − 0.30)(9.8) = 1.96 m/s^2.)
Maximum static force: F_max = μ_s N = 0.50(1127 N) = 563.5 N ≈ 5.64×10^2 N.
If you continue to push with that same force once it slips, kinetic friction is F_k = μ_k N = 0.30(1127 N) = 338.1 N.
Net force = F_max − F_k = 563.5 − 338.1 = 225.4 N.
Acceleration = a = (net force)/m = 225.4 N / 115 kg = 1.96 m/s^2 ≈ 2.0 m/s^2.
(You can also get a = (μ_s − μ_k) g = (0.50 − 0.30)(9.8) = 1.96 m/s^2.)
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