the end of the tracks, 8.8 m lower vertically, is a horizontally situated spring with constant 5 × 105 N/m. The acceleration of
1 answer:
Answer
Assuming the mass of the car, m = 43000 kg
initial speed u = 0
vertical distance moved, h = 8.8 m
spring constant k = 5 x 10⁵ N / m
acceleration of gravity = 9.8 m/s²
From law of conservation of energy ,
Gravitational potential energy at starting position =potential energy of the spring at maximum compression
x = 14.83 m
If the mass of the car is equal to 43000 Kg the spring is compressed to 14.83 m
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Answer:
The acceleration of the crate is 1.8 m/s² so the answer is a.
Explanation:
The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)
So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.
As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:
Starting with the sum of forces in the y-direction, we get:
Σ=0
We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.
so the sum will be:
when solving for N we get that:
N=W
where W is found by multiplying the mass of the crate by the acceleration of gravity:
N=250kg*9.8m/s²
N=2450N
Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:
μ
where μ is the kinetic friction coefficient.
So we get that the kinetic friction is:
so
With this information we can go ahead and find the sum of horizontal forces:
Σ=ma
In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.
so the sum of forces look like this:
750N-=ma
so
750N-294N=(250kg)a
when solving for a we get:
so the crate's acceleration is 1.82m/s².
Answer:
the period of the 16 m pendulum is twice the period of the 4 m pendulum
Explanation:
Recall that the period (T) of a pendulum of length (L) is defined as:
where "g" is the local acceleration of gravity.
SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:
therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.