In a real system of levers, wheel or pulleys, the AMA (actual mechanical advantage) is less than the IMA (ideal mechanical advantage) because of the presence of friction.
In fact, the IMA and the AMA of a machine are defined as the ratio between the output force (the load) and the input force (the effort):
however, the difference is that the IMA does not take into account the presence of frictions, while the AMA does. As a result, the output force in the AMA is less than the output force in the IMA (because some energy is dissipated due to friction), and the AMA is less than the IMA.
Answer:
a) Block 1 = 72.9kgm/s
Block 2 = 0kgm/s
b) vf = 1.31m/s
c) ∆KE = 936.36Joules
Explanation:
a) Momentum = mass× velocity
For block 1:
Momentum = 2.7×27
= 72.9kgm/s
For block 2:
Momentum = 53(0) (body is initially at rest)
= 0kgm/s
b) Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses of the block
u1 and u2 are their initial velocity
v is the common velocity
Given m1 = 2.7kg, u1 = 27m/s, m2 = 53kg, u2 = 0m/s (body at rest)
2.7(27)+53(0) = (2.7+53)v
72.9 = 55.7v
V = 72.9/55.7
Vf = 1.31m/s
c) kinetic energy = 1/2mv²
Kinetic energy of block 1 = 1/2×2.7(27)²
= 984.15Joules
Kinetic energy of block 2 before collision = 0kgm/s
Total KE before collision = 984.15Joules
Kinetic energy after collision = 1/2(2.7+53)1.31²
= 1/2×55.7×1.31²
= 47.79Joules
∆KE = 984.15-47.79
∆KE = 936.36Joules
Answer:
Explanation:
Given that,
Magnetic field of 0.24T
B = 0.24T
Field perpendicular to plane i.e 90°
Rate of decrease of length of side of square is 5.4cm/s
dL/dt = 5.4cm/s = 0.054m/s
Since it is decreasing
Then, dL/dt = -0.054m/s
When L is 14cm, what is the EMF induced?
L = 14cm = 0.14m
EMF is give as
ε = - dΦ/dt
Where flux is given as
Φ = BA
Where A is the area of the square
A = L²
Then, Φ = BL²
Substituting this into the EMF
ε = - dΦ/dt
ε = - d(BL²)/dt
B is constant
ε = - Bd(L²)/dt
ε = -2BL dL/dr
ε = -2 × 0.24 × 0.14 × -0.054
ε = 3.63 × 10^-3 V
ε = 3.63mV
Answer:
Change in translational KE is given as
Explanation:
As we know by work energy theorem that work done by all forces is equal to the change in kinetic energy of the system
So here we know that there are two forces acting on the system
So work done by the system of this force is equal to the change in kinetic energy
So we have
So we have
so we have
The maximum possible force of static friction between two surfaces is given by:
F = μN
F = max friction force, μ = coefficient of static friction, N = normal force
The normal force N equals the object's weight:
N = mg
m = mass, g = gravitational acceleration
Make a substitution:
F = μmg
Given values:
F = 500N, m = 100kg, g = 9.8m/s²
Plug in and solve for μ:
500 = μ(100)(9.8)
μ = 0.51
