The figure gives the acceleration of a 2.0 kg particle as an applied force moves
ID: 1788070 • Letter: T
Question
The figure gives the acceleration of a 2.0 kg particle as an applied force moves it from rest along an x axis from x = 0 to x-9.0 m. The scale of the figure's vertical axis is set by as 7.0 m/s. How much work has the force done on the particle when the particle reaches (a) x = 4.0 m, (b) x = 7.0 m, and (c) x = 9.0 m? what is the particle's speed and direction (give positive answer if the particle moves along x axis in positive direction and negative otherwise) of travel when it reaches (d) x = 4.0 m, (e) x = 7.0 m, and (f) x = 9.0 m? as x (m)Explanation / Answer
from the definiton of Work Dne W = F * s
W = ma * s
W = m * (a*s)
we get this term of a*s from as area under it .
so Work done = mass *area under graph
part A:
work done = m* area of trapezium
area of trapizium = area of traingle + area of rectangle
Area = (0.5* 1 * as) + ( length * breadth ) from x = 1 to x = 4
Area = 0.5 * 1* 7 + (4-1) * as)
Area = 3.5 + 3* 7
area = 24.5 m^2
W = m * area
W = 2*24.5 = 49 Joules
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(b)
Work for X= 4 m is 49 J
from x = 4 to x= 7 m
we have x = 4 to x=5 is +ve and x = 5 m to 6m is -ve .
so Work from x = 4 to x = 6 is 0
for x = 6 to 7 m, Work W = m* area from 6 to 7
W = 2 * *L*b)
W = 2*1 * -7 = -14 Joules
total work W = 49-14 = 35 Joules
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part C:
for x = 9m,
from 7m to 8m area = L *b = 1* 7 = -7 m^2
from 8 m to 9m , area = 0.5 *b h = 0.5* 1* -7 = -3.5 m^3
total area = -7-3.5 = -10.5 m^2
Work done W = 2*-10.5 = -21 Joules
total work DOne W = 35 -21 = 14 J
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part D :
Work Done W = KE = 0.5 mv^2
V^2 = 2 W/m
for x = 4m,
v^2 = 2* 49/2
v = 7 m/s
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part E:
for x = 7m,
Work Done W = KE = 0.5 mv^2
V^2 = 2 W/m
for x = 7m,
v^2 = 2* 35/2
v = 5.91 m/s
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part F:
for x = 9 m
Work Done W = KE = 0.5 mv^2
V^2 = 2 W/m
for x = 9m,
v^2 = 2* 14/2
v = 3.74 m/s
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