ept 15623598 A object of mass 3.00 kg is subject to a force Fo that varies with
ID: 1544589 • Letter: E
Question
ept 15623598 A object of mass 3.00 kg is subject to a force Fo that varies with position as in the figure below. F. (N) x (m) 2 4 6 8 10 12 14 16 18 20 a Find the work done by the force on the object asit moves from x- o to x m 5.00 m. Your response differs from the comect answer by more than 100%. J (b) Find the work done by the force on the object as it moves from x- 5.00 m to x 12.0 m Your response differs from the correct answer by more than 100%. J (c) Find the work done by the force on the object as it moves from x-12.0 m to x 17.0 m differs significantly from the correct answer. Rework your solution from the begin and check each step carefully. J (d) if the object has a speed of o 4so m/s at x 0, find its speed at x 5.00 m and its speed at x 17.o m. speed at x 5.00 m speed at x 17.0 m Need Help?Explanation / Answer
Given F vs displacement graph
a) we know taht work done W = F*s cos theta
and from the F vs s graph W = area under the curve
W from x = 0 to x=5 mis
W = area of triangle = 0.5*5*1.5 J = 3.75 J
b)
from x= 5 m to x= 12 m
work done = area under the curve
= 7*2 = 14 J
c) w from x= 12 to x= 17 m is
area of the triangle = 0.5*17*2 = 17 J
d) speed at x=0 is 0.450 m/s , at x= 5 m v=?
work done = dK.e = 0.5m(v2^2 -v1^2)
3.75 = 0.5*3(v^2 - 0.45^2)
v2 = 1.6439 m/s
at 17 m is W = w1+W2 +W3 =3.75+14+17 J= 34.75 J
now the velcoity at 17m is ,
34.75 = 0.5*3(v^2 - 0.45^2)
v2 = -4.8341 m/s
========================
a) from x=0 m to X = 16 m
work done = area under the curve
workdone = 4*6+4*6 = 48 J
b) from x = 16,o m and x = 24 m is
area under the curve is A = 2(0.5*2*6) Jc= -4.834 J
c)x= 0 to x=24 m n o is
W = 48 -4.834 J = 43.166 J
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