Answer:
Explanation:
Mass of 190kg
Coefficient of static friction is 0.4
Coefficient of kinetic friction 0.36
Horizontal force= 500N
Taking g=9.81m/s^2.
The weight of the body my
W=190×9.81=1863.91N
There is a normal acting on the body which is equal to the weight
N=W=1863.91N
Frictional force(fr) is acting on the body and it is opposite the horizontal force.
The minimum force to be overcome before the object can start to move is Fr = μsN
Fr= μsN. μs=0.4
Fr= 0.4×1863.91
Fr=745.56N.
Since the horizontal force (500N) is not up to the minimum force to make the object move, then the force of 500N the body is still at rest.
Then the frictional force at that time is equal to the horizontal force
Therefore
Functional force = 500N
b. Mass of asteroid is
M=2000kg
Asteroid velocity at a particular instant is,
U=(-1.30x10^4, 4.20x10^4, 0)m/s
Magnitude of U is
U=√(-1.30×10^4)^2 +(4.2×10^4)^2+0
U=√1.933E9
U=4.39×10^4m/s
Position of the asteroid from the centre of the earth is,
R= (6.00x10^6, 10.00x10^6, 0)m.
The magnitude of the radius is
R = √(6.00x10^6)^2+ (10.00x10^6)^2+ 0^2
R=√3.6E13+10E13+0
R=√13.6E13
R=1.17E7m
R^2=13.6E13m
The mass of the earth is
Me=5.97x10^24 kg
The momentum of the asteroid after time, t=1.5×10^3s
Given that G=6.67x10^-11Nm^2/kg^2
Momentum is
Mv-Mu=Ft
There the new momentum will be
Mv=Ft+Mu
Now we the to find the force the earth exert on the asteroid by using
F=GMMe/R^2
F=6.67E-11 ×2000× 5.97E24 /13.6E13
F=7.964E17/13.6E13
F=5855.88N
The new momentum
Mv= Mu+Ft
Mv= 2000(4.39E4)+5855.88(1.5E3)
Mv=9.66E7kgm/s
The new momentum is 9.66×10^7 Kgm/s
