You are designing an amusement park ride. A cart with two riders moves horizontally with speed v = 5.40 m/s . You assume that th
1 answer:
Answer:
Spring constant required=8910 N/m. Maximum distance compressed=0.99m
Explanation:
The first step here is to consider the Newton's Second Law which states that a force <em>F</em> applied to a body of mass <em>m</em> produces an acceleration <em>a</em>. Our body has a mass of 300 kg and the maximum allowable acceleration is 3.00 g. Considering gravity g=9.81m/s^2, <em>a</em>=3.00*9.81m/s^=29.43m/s^2. Hence,
F=300kg*29.43 m/s^2=8829 N
The force that a spring gives is the multiplication of its force constant <em>K </em>and the distance compressed<em> X.</em> . So the force F sholud be equal to <em>K*. </em>
<em>F=K*.</em>
Then, when the spring is fully compressed all the kinetic energy of the cart (T) is transferred to the spring as potential elastic energy (U). <em> </em>
<em> Eq.1</em>
We do not know but we do know that it is equal to F/<em>K. </em>Thus,
Eq.2
Operating in Eq.2
Finally, putting the valus of <em>K </em>in Eq.1 and solving, <em>=0.99m</em>
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(a) No, because the mechanical energy is not conserved
Explanation:
The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:
(1)
However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.
Therefore, eq. (1) can be rewritten as
which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane () and part is lost because of the air resistance (
).
(b) 77.8 m/s
First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:
Now we can calculate the acceleration of the plane, by using Newton's second law:
where m is the mass of the plane.
Finally, we can calculate the final speed of the plane by using the equation:
where
is the final velocity
is the initial velocity
is the acceleration
is the distance travelled
Solving for v, we find
Answer:
Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is
The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is
The friction forces are computed by
Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]
Simplifying
Plugging in the values
