Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached
to an eye anchor embedded in a wall. Rope segment BC creates an angle of ϕ = 51.0 ∘ with the floor and rope segment CD creates an angle θ with the horizontal. If both ropes BCA and CD can support a maximum tensile force Tmax = 120 lb , what is the maximum weight Wmax of the crate that the system can support? What is the
1 answer:
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Answer:
Δ KE = 249158.6 kJ
Explanation:
given data
Truck mass M = 1560 Kg
Truck initial speed, u = 28 m/s
mass of car m = 1070 Kg
initial speed of car u1 = 0 m/s
solution
first we get here final speed by using conservation of momentum that is express as
Mu = (M+m) V .......................1
put here value we get
1560 × 28 = (1560 + 1070 ) V
solve it we get
final speed V = 16.60 m/s
and
Change in kinetic energy will be here
Δ KE = .................2
put here value and we get
Δ KE =
solve it we get
Δ KE = 249158.6 kJ
Answer:
A: The acceleration is 7.7 m/s up the inclined plane.
B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane
Explanation:
Let us work with variables and set
As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.
Part A:
From the free body diagram we see that the total force along the x-axis is:
Now the force of friction is where
is the normal force and from the diagram it is
Thus
Therefore,
Substituting the value for we get:
Now acceleration is simply
The negative sign indicates that the acceleration is directed up the incline.
Part B:
Which can be rearranged to solve for t:
Substitute the value of and
and we get:
which is our answer.
Notice that in using the formula to calculate time we used the positive value of , because for this formula absolute value is needed.
