Starting from rest, a 12-kg cardboard box containing used textbooks slides down a steel ramp that is 1.5 m high and 4.5 m long. At the end of the ramp the block is moving at 3.6 m/s.
What is the force of friction between the box and the ramp?
Starting from rest, a 12-kg cardboard box containing used textbooks slides down a steel ramp...?
2020-09-30 12:47 am
回答 (1)
2020-09-30 1:22 am
Method 1 :
Refer to the figure below.
Take g = 9.8 m/s²
tanθ = 1.5/4.5
θ = 18.44°
Consider the sliding down of the cardboard along the ramp :
Initial velocity, u = 0 m/s
Final velocity, v = 3.6 m/s
Displacement, s = 1.5/sin18.44°
v² = u² + 2as
3.6² = 0² + 2a(1.5/sin18.44°)
Acceleration, a = 1.37 m/s²
F = ma
mg sinθ - f = ma
(12)(9.8)(sin18.44°) - f = (12)(1.37)
Frictional force, f = 21 N
====
Method 2 :
Refer to the figure below.
Take g = 9.8 m/s²
Decrease in P.E. = Increase in K.E. + Work done against friction
mgh = (1/2)mv² + fs
(12)(9.8)(1.5) = (1/2)(12)(3.6)² + f(1.5/sin18.44°)
f = [(12)(9.8)(1.5) - (1/2)(12)(3.6)²] × sin18.44°/1.5
f = 21 N
Refer to the figure below.
Take g = 9.8 m/s²
tanθ = 1.5/4.5
θ = 18.44°
Consider the sliding down of the cardboard along the ramp :
Initial velocity, u = 0 m/s
Final velocity, v = 3.6 m/s
Displacement, s = 1.5/sin18.44°
v² = u² + 2as
3.6² = 0² + 2a(1.5/sin18.44°)
Acceleration, a = 1.37 m/s²
F = ma
mg sinθ - f = ma
(12)(9.8)(sin18.44°) - f = (12)(1.37)
Frictional force, f = 21 N
====
Method 2 :
Refer to the figure below.
Take g = 9.8 m/s²
Decrease in P.E. = Increase in K.E. + Work done against friction
mgh = (1/2)mv² + fs
(12)(9.8)(1.5) = (1/2)(12)(3.6)² + f(1.5/sin18.44°)
f = [(12)(9.8)(1.5) - (1/2)(12)(3.6)²] × sin18.44°/1.5
f = 21 N
收錄日期: 2021-04-30 17:51:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200929164739AA2Ssqg