A box moving at 2m/s slides to a stop over 2 m. If the box has a mass of 40kg, what is the value of μ
between box and the floor? Use g = 10m/s^2
Physics homework?
2019-02-20 11:08 am
回答 (2)
2019-02-20 11:26 am
Method 1 :
Work done against frictional force = Lost in K.E.
f × (2 m) = (1/2) × (40 kg) × (2 m/s)²
Frictional force, f = 20 N
Normal force, N = mg = (40 kg) × (10 m/s²) = 400 N
Friction coefficient, μ = f/N = (20 N) / (400 N) = 0.05
Method 2 :
v² = vₒ² + 2as
(0 m/s)² = (2 m/s)² + 2a(2 m)
Acceleration, a = -0.5 m/s²
Frictional force, f = ma = (40 kg) × (0.5 m/s²) = 20 N
Normal force, N = mg = (40 kg) × (10 m/s²) = 400 N
Friction coefficient, μ = f/N = (20 N) / (400 N) = 0.05
Work done against frictional force = Lost in K.E.
f × (2 m) = (1/2) × (40 kg) × (2 m/s)²
Frictional force, f = 20 N
Normal force, N = mg = (40 kg) × (10 m/s²) = 400 N
Friction coefficient, μ = f/N = (20 N) / (400 N) = 0.05
Method 2 :
v² = vₒ² + 2as
(0 m/s)² = (2 m/s)² + 2a(2 m)
Acceleration, a = -0.5 m/s²
Frictional force, f = ma = (40 kg) × (0.5 m/s²) = 20 N
Normal force, N = mg = (40 kg) × (10 m/s²) = 400 N
Friction coefficient, μ = f/N = (20 N) / (400 N) = 0.05
2019-02-20 11:26 am
Let’s use the following equation to determine the acceleration of the box.
vf^2 = vi^2 + 2 * a * d
0 = 2^2 + 2 * a * 2
a = -1 m/s^2
a = -μ * 10
-μ * 10 = -1
μ = 0.1
vf^2 = vi^2 + 2 * a * d
0 = 2^2 + 2 * a * 2
a = -1 m/s^2
a = -μ * 10
-μ * 10 = -1
μ = 0.1
收錄日期: 2021-04-24 01:16:55
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