A 7.00 kg mass is being pulled by a 53.8 n force what is the acceleration
1 answer:
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Answer:
A. A = 0.913 m
B. amax = 132.24m/s^2
C. Fmax = 324.01N
Explanation:
When the block is moving at the equilibrium point , its velocity is maximum.
A. To find the amplitude of the motion you use the following formula for the maximum velocity:
(1)
vmax = maximum velocity = 11.0 m/s
A: amplitude of the motion = ?
w: angular frequency = ?
Then, you have to calculate the angular frequency of the motion, by using the following formula:
(2)
k: spring constant = 355 N/m
m: mass of the object = 2.54 kg
Next, you solve the equation (1) for A and replace the values of vmax and w:
The amplitude of the motion is 0.913m
B. The maximum acceleration of the block is given by:
The maximum acceleration is 132.24 m/s^2
C. The maximum force is calculated by using the second Newton law and the maximum acceleration:
It is also possible to calculate the maximum force by using:
Fmax = k*A = (355N/m)(0.913m) = 324.01N
The maximum force exertedbu the spring on the object is 324.01 N