The net force on the block perpendicular to the floor is
∑ F[perp] = F[normal] - mg = 0
so that
F[normal] = (5 kg) g = 49 N
Then
F[friction] = 0.1 F[normal] = 4.9 N
so that the net force parallel to the floor is
∑ F[para] = -4.9 N = (5 kg) a
Solve for the acceleration a :
a = (-4.9 N) / (5 kg) = -0.98 m/s²
Starting with an initial velocity of 5 m/s, the box comes to a stop after time t such that
0 = 5 m/s - (0.98 m/s²) t
⇒ t ≈ 5.1 s
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Answer: 4.98 m/s
Explanation:
You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.
PEi = 0 in this case
KEi = ½mVi² = PEf+KEf = mghf + ½mVf²
½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²
½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26
Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s
Answer:
2.068 x 10^6 m / s
Explanation:
radius, r = 5.92 x 10^-11 m
mass of electron, m = 9.1 x 10^-31 kg
charge of electron, q = 1.6 x 10^-19 C
As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.
centripetal force = 
Electrostatic force = 
where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2
So, balancing both the forces we get
v = 2.068 x 10^6 m / s
Thus, the speed of the electron is give by 2.068 x 10^6 m / s.
Answer:
A) When the angle between the Force (F) and Displacement (x) is 0°, because, Work done (W) is directly proportional to the Cosine of the Angle between the Force applied and the resultant displacement of the subject.
W = F•x cos ∅
If ∅ = 0°,
W = F•x ===> Maximum Work Done.
If ∅ = 45°,
W = F•x/√2
If ∅ = 90°,
W = 0
If ∅ = 180°,
W = –F•x ===> Minimum Work Done.
