How large is the force of friction over the box ?
2017-06-07 12:25 am
A 12 kg box is slowed by friction from 8 m/s to 3 m/s over 12.5 m. How large is the force of friction on the box? ** please explain so I understand **
回答 (1)
2017-06-07 12:40 am
Method 1 :
Initial velocity, u = 8 m/s
Final velocity, v = 3 m/s
Displacement, s = 12.5 m
v² = u² + 2as
3² = 8² + 2×a×12.5
Acceleration, a = -2.2 m/s²
f = ma
Fictional force, f = 12 × (2.2) = 26.4 N
====
Method 2 :
Loss in K.E. in J = (1/2) × 12 × 8² - (1/2) × 12 × 3²
Work done by friction in J = f s = f × 12.5
Work done by fiction = Loss in K.E.
f × 12.5 = (1/2) × 12 × 8² - (1/2) × 12 × 3²
f × 12.5 = 330
Fictional force, f = 330/12.5 = 26.4 N
Initial velocity, u = 8 m/s
Final velocity, v = 3 m/s
Displacement, s = 12.5 m
v² = u² + 2as
3² = 8² + 2×a×12.5
Acceleration, a = -2.2 m/s²
f = ma
Fictional force, f = 12 × (2.2) = 26.4 N
====
Method 2 :
Loss in K.E. in J = (1/2) × 12 × 8² - (1/2) × 12 × 3²
Work done by friction in J = f s = f × 12.5
Work done by fiction = Loss in K.E.
f × 12.5 = (1/2) × 12 × 8² - (1/2) × 12 × 3²
f × 12.5 = 330
Fictional force, f = 330/12.5 = 26.4 N
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