2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is
applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?
2 answers:
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(a) See graph in attachment
The appropriate graph to draw in this part is a graph of velocity vs time.
In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.
Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at
t = 10 s
v = 15 m/s
(b)
The average acceleration of the horse can be calculated as:
where
v is the final velocity
u is the initial velocity
t is the time interval
In this problem,
u = 0
v = 15 m/s
t = 10 s
Substituting,
(c) 75 m
For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:
where
s is the distance travelled
u is the initial velocity
t is the time interval
a is the acceleration
In this problem,
u = 0
t = 10 s
Substituting,
(d) See attached graphs
In a uniformly accelerated motion:
- The distance travelled (x) follows the equation mentioned in part c,
So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.
- The acceleration is constant during the motion, and its value is
(calculated in part b)
therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.
Answer:
= 285 Joules
Explanation:
a) answer can be found out in attachment
(b) The temperature for the isothermal compression is the same as the temp at the end of the isobaric expansion. Since pressure is held constant but volume doubles, we use the ideal gas law:
p V = nR T to see that the temperature also doubles.
.So... temp for isothermal compression = 355×2 = 710 K
.(c) The max pressure occurs at the top point. At this point, the volume is back to the original value but the temperature is twice the original value. So the pressure at this point is twice the original, or
max pressure = 2×240000 Pa = 480000 Pa = 4.80 x 10^5 Pa
(d) total work done by the piston = workdone during isothermal compression - work done during expansion =
= nRT ln(V initial / V final)-p (V initial - V final)
= nRT ln(2) - nR(T final - T initial)
= 0.250× 8.314 ×710×ln(2)-0.250×8.314× (710 - 355)
= 285 Joules
