wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An engine of mass 5×104 kg pulls a coach of mass 4×104 kg. A resistance of 1 N per 100 kg is acting on both the coach and the engine and if the driving force of the engine is 4500 N then the acceleration of the engine and the tension in the coupling will respectively be_________

A
The acceleration of the system (coach+engine) is 0.04m/s2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
The acceleration of the system (coach+engine) is 0.05m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
The tension in the coupling is 2000 N.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
The tension in the coupling is 2400 N.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct options are
A The acceleration of the system (coach+engine) is 0.04m/s2
C The tension in the coupling is 2000 N.
Drivingforce=4500 N

Opposing force Resistance f
f=(5+4)×104100
f=900N
Resultant force:
FR=4500900
FR=3600 N
Mass of engine and coach =9×104 kg

According to Newton's law
F=ma
3600=9×104×a
a=36009×104
a=0.04m/s2

Now considering the eqilibrium of the coach only
We have,
(TR)=4×104×0.04(F=ma)
T4×104100=4×104×0.04
T=(4×104×0.04) + (4×102)
T=1600+400
T=2000 N

Hence, Option A and C are answer.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Second Law of Motion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon