A 4.6 kg block is slid along the floor through different paths as shown above. T
ID: 1628610 • Letter: A
Question
A 4.6 kg block is slid along the floor through different paths as shown above. The co-efficient of kinetic friction between the block and the floor is 0.24.
1)What is the work down by friction when the block moves from point A to point B along back 3?
2)What is the work done by friction if the block moves in a closed loop, starting at point A along path 2 and then back along path 3?
3)What is the work done by friction if the blcok moves in a closed loop, starting at point A along path 1 and then back along path 2?
4)) What is the work done by friction if the block moves in a closed loop, starting at point A along path 1 and then returning along the same path?
4.0m 1.0m 10m 1.0m 10m 1.0m 4.0m 2) 8 2.0m 3.0m 2.0m 1.0m 3.0m3 3 0mExplanation / Answer
Given,
Mass of block, m = 4.6 kg
Coefficient of friction, u = 0.24
frictional force, Ff = 0.24 x 4.6 x 9.8 = 10.82 N
Work = 10.82 x distance (d)
1) Work down by friction when the block moves from point A to point B along back 3
For path 3, d = 1 + 3 + 3 = 7 m
So, WFric-3 = 10.82 x 7 =75.73 Nm
2) Work done by friction if the block moves in a closed loop, starting at point A along path 2 and then back along path 3
For path 2, d = 2 + 2 + 1 = 5 m
For path 2 and 3, d = 12 m
WFric-23 = 10.82 x 12 = 129.84 Nm
3) Work done by friction if the blcok moves in a closed loop, starting at point A along path 1 and then back along path 2
For path 1, d = 4 + 4 + 1 + 1 + 1 = 11 m
For path 1 and 2, d = 11 + 5 =16
WFric-12 = 10.82 x 16 = 173.12 Nm
4) Work done by friction if the block moves in a closed loop, starting at point A along path 1 and then returning along the same path
d = 11 + 11 = 22
WFric-11 = 10.82 x 22 = 238.04 Nm
5) Work done by a conservative force if the block moves in a closed loop, starting at point A along path 1 and then back along path 2
If we assume that the block is moving at a constant speed, the conservative force is equal to the friction force. So, the answer to this question is the same as the answer to part 3. i.e.
Wconserv-12 = 10.82 x 16 = 173.12 Nm
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