Jeff’s father is installing a do-it-yourself security system at his house. He needs to get a device from his workshop that conve
2 answers:
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Answer:
a) Final position is x = 0.90 m
b) Final position is x = 0.133 m
Explanation:
The workdone between two points is usually approximated as the area under the force-distance curve between those two points.
From the graph,
As at the initial position, x = 0.40 m and the corresponding F = 0.8 N,
The area from that point onwards up to the end of that particular bar = 0.8 (0.5 - 0.4) = 0.08 J
The next bar has force = 0.4 N and the width of the bar = (0.75 - 0.50) = 0.25 m
Work done under this bar = 0.4 × 0.25 = 0.1 J
Total work done from the starting position up to this point now = 0.08 + 0.1 = 0.18 J, still less than 0.21 J
So, the final position has to be on the last bar. Let the position be x. The force on the last bar = 0.2 N
0.21 = 0.18 + 0.2 (x - 0.75)
0.03 = 0.2x - 0.15
0.2x = 0.18
x = 0.9 m
Therefore, the final position of the object, to do 0.21 J worth of work, starting from x = 0.4 m is 0.90 m.
b) For this part, negative work is done, this means, we will move in the negative direction to try and trace this total work done.
From the starting point where the initial position is 0.40 m, the force here is 0.80 N
The workdone under this bar to the left is
The workdone = 0.8 (0.25 - 0.4) = - 0.12 J
Since we're tracing -0.19 J, the final position has to be on the last bar (on the left), Let the position be x. The force on the last bar on the left (could also be referred to as the first bar) = 0.60 N
- 0.19 = -0.12 + 0.6 (x - 0.25)
-0.07 = 0.6x - 0.15
0.6x = 0.08
x = (0.08/0.6) = 0.133 m
Therefore, the final position of the object, after doing -0.19 J worth of work, starting from x = 0.4 m is 0.133 m.
Hope this Helps!!!
Answer:
The time interval of acceleration for the bus is 2.20 seconds
Explanation:
Acceleration is the rate of change of velocity
→
where a is the acceleration, v is the final velocity, u is the initial velocity
and t is the time
The given is:
The uniform acceleration = -4.1 m/s²
The bus slows from 9 m/s to 0 m/s
We need to find the time interval of acceleration for the bus
Lets use the rule above
→ a = -4.1 m/s² , v = 0 m/s , u = 9 m/s
→
Multiply both sides by t
→ -4.1 t = -9
Divide both sides by -4.1
∴ t = 2.20 seconds
<em>The time interval of acceleration for the bus is 2.20 seconds</em>