A block of mass m slides on a horizontal frictionless surface. The block is attached to a spring with a spring constant K. At th
e instant shown below, the spring is stretched and the block is traveling to the right. Neglect air resistance. Equilibrium position
a) Draw the free body diagram for the block.
b) Apply Newton's Second Law, Fnetma, to the block for the x direction. Put into your equation the fact that the spring force magnitude, Fopring Kx, and solve for the acceleration of the bloc.
c) Start with the position equation x x cos( t+%) and differentiate twice to get dt d2x d2x
d)Sct a% from part b) equal to dtx from part c)toshow that ω-ν e) Write the frequency f in terms of K and m. f Write the period T in terms of K and m. g) Write the block velocity V as a function of time. h) Write the block acceleration 'a' as a function of time.
a) Draw the free body diagram for the block.
b) Apply Newton's Second Law, Fnetma, to the block for the x direction. Put into your equation the fact that the spring force magnitude, Fopring Kx, and solve for the acceleration of the bloc.
c) Start with the position equation x x cos( t+%) and differentiate twice to get dt d2x d2x
d)Sct a% from part b) equal to dtx from part c)toshow that ω-ν e) Write the frequency f in terms of K and m. f Write the period T in terms of K and m. g) Write the block velocity V as a function of time. h) Write the block acceleration 'a' as a function of time.
1 answer:
Answer:
b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e) T = 2π √m / k
h) a = - A w² cos (wt+Ф)
Explanation:
a) see free body diagram in the attachment
b) We write Newton's second law
Fe = m a
-k x = ma
a = -k / m x
c) the acceleration is
a = d²x / dt²
If x = A cos wt
v = dx / dt = -A w sin (wt +Ф)
a = d²x / dt² = - A w² cos (wt+Ф)
d) we substitute in Newton's second law
d²x / dt² = -k / m x
We call
w² = k / m
e) substitute to find w
-A w² cos (wt+Ф) = -k / m A cos (wt+Ф)
w² = k / m
Angular velocity and frequency are related
w = 2π f
f = 1 / T
We substitute
T = 2π / w
T = 2π √m / k
g) v= - A w sin (wt+Ф)
h) acceleration is
a = - A w² cos (wt+Ф)
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Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
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