01:09:00
CAD-01764 6.5 2.5 E E 5.5 2.5 average 4.45 1.25 D D 2.25 0.25 C C 0.3 0.04 average 0.35 0.1 B B 0.2 0.1 A A 0.05 0.001 average 0.1 0.03
F F 5.3 1.25 G G 1.35 0.15 H H 0.55 0.05 average 1.5 0.15
I I 10.5 2.75 J J 3.3 0.15 K K 4.25 0.25 average 5.5 1.25
L L 2.95 0.55 M M 4.15 0.15 N N 3.2 0.1 average 3.5 0.25
O O 3.1 0.45 P P 3.2 0.2 Q Q 3.8 0.3 average 3.5 0.3
R R 4.1 0.45 S S 3.5 0.15 T T 1.05 0.05 average 2.75 0.25
U U 3.5 0.15 V V 2.75 0.05 W W 4.35 0.25 average 3.5 0.15
X X 3.5 0.25 Y Y 4.5 0.15 Z Z 3.95 0.65 average 3.9 0.3
The measured values in the table are adjusted to make the total force equal to 8.0 N, so the values are decreased by 12.5%. The values must be multiplied by 0.875 to be corrected.
To verify the correctness of this table, we calculate the total force:
0.935 + 0.885 + 0.805 + 0.715 + 0.61 + 0.51 + 0.38 + 0.285 + 0.155 + 0.045 + 0.45 + 0.715 + 0.885 + 0.545 + 0.4 + 0.45 + 0.655 + 0.8 + 0.885 + 0.91 + 0.85 + 0.73 + 0.52 + 0.415 + 0.285 = 8.0 N
This proves that the table is correct.
## Related Questions
Mike throws 10.0 kg of Peach upon his sister. He does the throw and pushes 15.0 m. Knowing that the coefficient of friction and Peach is 0.15, what is the force of friction that Peach makes upon the ground?
Given data:
- Mass of Peach (m) = 10.0 kg
- Coefficient of friction (μ) = 0.15
- Gravitational acceleration (g) = 9.81 m/s²
1. **Calculate the normal force (N):**
The normal force is equal to the gravitational force acting on the Peach.
[
N = m imes g
]
[
N = 10.0 , ext{kg} imes 9.81 , ext{m/s²} = 98.1 , ext{N}
]
2. **Calculate the force of friction (F):**
The force of friction is given by:
[
F = mu imes N
]
[
F = 0.15 imes 98.1 , ext{N} = 14.715 , ext{N}
]
Therefore, the force of friction that Peach makes upon the ground is **14.7 N**.
5月1日2014年
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