Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
Force acts for 20 s on a body of mass 20 kg, starting from rest, after which the force ceases and then body describes 50 m in the next 10 s. The value of force will be :
Question Error Report
Question is incomplete/wrong
Question not belongs to this Chapter
Answer is wrong
Solution is wrong
Answer & Solution is not matching
Spelling mistake
Image missing
Website not working properly
Other (not listed above)
Error description
Thank you for reporting, we will resolve it shortly
Back to Question
Thank you for reporting, we will resolve it shortly
Q. Force acts for $20 \,s$ on a body of mass $20\, kg$, starting from rest, after which the force ceases and then body describes $50\, m$ in the next $10 \,s$. The value of force will be :
JEE Main
JEE Main 2023
Laws of Motion
A
40 N
6%
B
5 N
43%
C
20 N
32%
D
10 N
18%
Solution:
$ 50=V \times 10$
$V=5 \,m / s $
$ V =0+ a \times 20$
$ 5= a \times 20 $
$ a =\frac{1}{4} \,m / s ^2 $
$ F = ma =20 \times \frac{1}{4}=5 N $