A .15-kg object moves along a straight line. the net force acting on the object
ID: 2254618 • Letter: A
Question
A .15-kg object moves along a straight line. the net force acting on the object varies with the object's displacement as shown in the graph below. the object starts from rest at displacement x=0 and time t=0 and is displaced a distance of 20m. determine each of the following:
a) the acceleration of the particle when its dispacement x=6m
b) the time taken for the object to be displaced for the first 12m
C) the amount of work done by the net force displacing the object the first 12m
d) the speed of the object at displacement x=12m
e) the final speed at displacement x=20m
f) the change in momentum of the object as it is diplaced from x=12m to x=20m
g)the impulse of the object recieved during the total displacement.
Please show all your work! the picture wont upload, but i can email it out to you.
Explanation / Answer
1)
f=ma
a=f/m = 4/0.15 = 26.666 m/s^2
2)
x=0.5a*t*t
t= sqrt(2x/a)
=sqrt( 2*12/26.666) = 0.948 seconds
3)
it is the area under the graph for the first 12 sec = 4*12 = 48 Joules
4)
v*v = u*u + 2*a*s
v*v = 0*0 + 2*26.666*12
v = 25.2979 m/s
5)
we can use the conservation of energy here
total energy = area under the whole curve = 48 + 0.5 * 8 * 4 = 64 Joules
final kinetic energy = 0.5 * m*v*v
64 = 0.5*0.15*v*v
v = 29.2118 m/s
6)
The change in the momentum of the object as it is displaced from x = 12 m to x = 20 m
= m(v-u)
= 0.15 * (29.2118 - 25.2979)
=0.587085 Kg-m/s
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