Question : A body of mass 0.40 kg moving initially with a constant speed of 10 m/s to the north is subjected to a constant force of 8.0 N directed towards the south for 30s. Take the instant the force is applied to be t=0, the position of the body at that time to be x=0, and predict its position at t=-5s, 25s, 100s.
Doubt by Krish
Solution :
Mass of the body (m) = 0.40 kg
Initial speed of the body (u) = +10 m/s (North)
Force applied on the body (F) = -8N (South)
Duration of force applied (t) = 30 sec
At t=0, x=0
Before the application of applied force, the body is moving with a uniform velocity of 10 m/s and after the duration of 30 seconds again the body keeps moving with a velocity of 10 m/s.
i) Position of the body at t=-5sec.
acceleration of the body (a)=0 m/s²
initial velocity of the body (u)=10 m/s
Using
Using
s=ut+½at²
s1=(10)(-5)+½(0)(-5)²
s1=-50+0
s1=-50 m
s1=-50 m
Hence, at -5sec, the position of the body will be -50 m. Negative sign shows that the body will be 50 m towards south.
ii) Position of the body at t=25sec.
acceleration of the body (a)=F/m
= -8/0.40
= -20 m/s²
= -8/0.40
= -20 m/s²
initial velocity of the body (u)=10 m/s
Using
Using
s=ut+½at²
s2=10(25)+½(-20)(25)²
s2=250-6250
s2=-6000 m
Hence, at 25sec,the position of the body will be -6000 m. Negative sign shows that the body will be 6000 m towards south.
iii) Position of the body at t=100 sec.
In this case, there will be two different cases :
Distance covered by the body from t=0 to t=30s & Distance covered by the body from t=30s to t=100s.
a) Distance covered by the body from t=0 to t=30s
Here
Initial velocity (u) = 10 m/s
acceleration of the body (a) = F/m
acceleration of the body (a) = F/m
= -8/0.40
= -20 m/s²
= -20 m/s²
Using
s=ut+½at²
s=(10)(30)+½(-20)(30)²
s=300-10×900
s=300-9000
s=300-10×900
s=300-9000
s=-8700 m
b) Distance covered by the body from t=30 to t=100s
Here
Initial velocity (u') = Final velocity at the end of 30 seconds.
Using
v=u+at
u'=10+(30)(-20)
u'=10-600
u'=-590 m/s
Using
v=u+at
u'=10+(30)(-20)
u'=10-600
u'=-590 m/s
acceleration of the body (a) = 0 m/s²
[∵ Force is acting up to 30 seconds only]
Using
s=ut+½at²
s'=u't+½at²
s'=(-590)(100-30)+½(0)t²
s'=-590×70+0
s'=-41300 m
s'=(-590)(100-30)+½(0)t²
s'=-590×70+0
s'=-41300 m
Total distance covered by the body in 100s
s3=s+s'
s3=-8700+(-41300)
s3=-8700+(-41300)
s3=-8700-41300
s3=-50000 m
Hence, at 30sec,the position of the body will be -50000 m. Negative sign shows that the body will be 50000 m towards south.