A 0.20-kg object moves along a straight line. The net force acting on the object

Question

A 0.20-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 above. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. The acceleration of the particle when its displacement x is 6 m The time taken for the object to be displaced the first 12 m The amount of work done by the net force in displacing the object the first 12 m The speed of the object at displacement x = 12 m The final speed of the object at displacement x = 20 m The change in the momentum of the object as it is displaced from x = 12 m to x = 20 m. Determine each of the following:

a) The acceleration of the particle when its displacement x is 6 m

b) The time taken for the object to be displaced the first 12 m

c) The amount of work done by the net force in displacing the object the first 12 m

d) The speed of the object at displacement x = 12 m

e) The final speed of the object at displacement x = 20 m

f) The change in the momentum of the object as it is displaced from x = 12 m to x = 20 m

Explanation / Answer

Given data
mass of the object = 0.20 kg
distance = 20 m
a)F=ma
a=F/m
a = 4/0.20=20 m/s^2

b)
x=1/2*a*t^2
t = sqrt(2*x/a)
t = sqrt(2*12/20)=1.10s
c)The amount of work done by the net force in displacing the object the first 12 m
work = are under the curve = 4*12=48 J
d) the speed of the object at displacement x=12 m
v^2 = v0^2+2*a*x
v=sqrt(2*a*x) = sqrt(2*20*12)=22 m/s
e)the final speed of the object at displacement x=20m
K=1/2 mv^2 = total area under curve=4*1+1/2*4*(20-12)=64 J
K=1/2*m*v^2
v=sqrt(2*K/m)=sqrt(2*64/0.20)=25 m/s
f)The change in the momentum of the object as it is displaced from x=12m to x= 20m
delta P = m*(v-v0)=0.20*(25-22)=0.60 kg.m/s

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