At 6: 00 am, a motorbike set off from town A to town B at a speed of 40km/h. At the same time, a car set off from town B to town
1 answer:
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Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
12) The weight of the plane is
13) The lift provided by the wing is
Explanation:
12)
The weight of an object is equal to the force of gravity acting on the object. Near the Earth's surface, it can be calculated as
where
W is the weight
m is the mass of the object
g is the acceleration of gravity
In this problem, we know that the mass of the plane is
m = 75 tonnes
Since 1 ton = 1000 kg, the mass in kg is
m = 75,000 kg
Also, the acceleration of gravity is
Therefore, the weight of the plane is:
13)
We can solve this question by applying Newton's second law along the vertical direction of motion. In fact, the net force acting along the vertical direction must be equal to the product between the mass of the plane and the vertical acceleration:
where
is the net force in the y-direction
m is the mass of the plane
is the acceleration in the y-direction
There are two forces acting in the vertical direction:
- The lift L, acting upward
- The weight W, acting downward
So the vertical net force is
And the equation becomes
Also, we know that the plane is travelling in level flight, which means that the vertical acceleration is zero:
Therefore, we get
And so the lift is
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