1) given data:
mass of block (m) = 2 kg
final velocity of block (V') = 10 m/s
distance travelled (d) = 20 m
coefficient of kinetic friction (
)
= 0.2
let the initial velocity of block be V

friction force =
N
=
*mg
= 0.2*2*9.81
= 3.924 N
from netwon's second law
f = m*a
a = f/m
= 3.924/2
= 1.962 m/s2
now from equation of motion
V'2 = V2 - 2*a*d
102 + 2*1.962*20 = V2
V = 13.359 m/s
thus correct option is c
2)
thus correct option is b
Question 2 A 2-kg block slides along a rough horizontal surface and slows to 10 m/s...
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