A car is moving with a speed of 15 m/s on a straight road is brought to rest while covering a
3
distance of km. If mass of car is 600 kg. Find the retarding force on the car.
20
(A) 200 N
(B) 225 N
(C) 450 N
(D) none of these
be
Share
A car is moving with a speed of 15 m/s on a straight road is brought to rest while covering a
3
distance of km. If mass of car is 600 kg. Find the retarding force on the car.
20
(A) 200 N
(B) 225 N
(C) 450 N
(D) none of these
be
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
Answer:
Answer:
C) 450 N
Explanation:
Initial velocity(u)=15 m/sec
Car is brought to rest so it's final velocity(u)=0 m/sec
Distance travelled(s)=3/20 km=150 m
Applying v^2 = u^2 + 2as
0^2 = 15^2 + 2 × a × 150
-225=300a
-225/300=a
a=-225/300
Mass of car=600 kg
Force=ma
Force=600 × -225/300
=2 × -225
=-450 N
Wasn't this a 8th class big bang edge fiitjee question?
I also went to give it..i'm in 8th too
Explanation:
Answer:
U is 15 m/s
V is 0m/s because at last the object comes at rest.
S is 3/20km so we convert it into m/s .on converting it is 3/20×1000 m per second .
using v^2 - u^2=2as
a=3/4 m/s^2( by calculations )
using F=M×A
F=600Kg × 3/4m/s^2
F=450N
Answer:
d) None of these
Explanation:
u=15m/s
v=0
s=distance travelled=3/20km
By newtons third law of motion,
v^2-u^2=2as
0-15^2=2×a×3/20
a=-225×10/3 m/s.
RETARDING FORCE=MASS×DECELERATION
= 600×-225×10/3
=-450000N
If you find any answer helpful mark me as brainliest