A 50kg object accelerates a 5m/s/s. What is the net force acting on the object?
2 answers:
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Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
Answer:
a)
b)
Explanation:
Given:
mass of bullet,
compression of the spring,
force required for the given compression,
(a)
We know
where:
a= acceleration
we have:
initial velocity,
Using the eq. of motion:
where:
v= final velocity after the separation of spring with the bullet.
(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,
∵At maximum height the final velocity will be zero
Using the equation of motion:
where:
h= height
g= acceleration due to gravity
is the height from the release position of the spring.
So, the height from the latched position be: