Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151
Answer:
Explanation:
Given that
F=ax^3/2. a is a constant
The force does a work of
W=2.01KJ from x=0 to x=15.2m
We need to find a
Work is give as,
W=∫F.ds
But this is in x direction only then,
W=∫Fdx. from x=0 to x=15.2m
W=∫ax^3/2dx from x=0 to x=15.2m
W=ax^(3/2+1)/(3/2+1).
W=ax^(5/2)/5/2
W=ax^(2/5)/2.5 from x=0 to x=15.2m
Cross multiply
2.5W=ax^2.5. from x=0 to x=15.2m
2.5W= a (15.2^2.5-0)
W=2.01KJ=2010J
2.5×2010=a×900.76
Therefore,
a=5.56
First we can say that since there is no external force on this system so momentum is always conserved.
now by the condition of elastic collision
![v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6](https://tex.z-dn.net/?f=v_%7B2f%7D%20-%20v_%7B1f%7D%20%3D%200.8%20-%200%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Enow%20add%20two%20equations%3C%2Fp%3E%3Cp%3E%5Btex%5D3%2Av_%7B2f%7D%20%3D%201.6)
also from above equation we have
So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.
Answer:
wrong answer because m=100
