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The magnitude of the average friction force exerted on the collar (F)= 8.641 N
<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>To calculate the magnitude of the average friction force exerted on the collar we are using the formula,
Here we are given,
k = The spring has a spring constant.
= 25.5 N/m.
= Final length of the spring .
=
= 1.591 m
= The initial length of the spring.
= 1.25−0.60
=0.65 m
y=The collar then travels downward a distance.
= 1.80 m.
m= The mass of the collar.
=3.55 kg
= the velocity of the collar.
= 3.39 m/s.
g = The acceleration due to gravity.
= 9.81 m/s²
We have to calculate the magnitude of the average friction force exerted on the collar = F
Now we put the known values in the above equation, we get;
Or,
Or, F= 8.641 N
From the above calculation we can conclude that,
The magnitude of the average friction force exerted on the collar (F)= 8.641 N
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Disclaimer: This question is incomplete in the portal. Here is the complete question.
Question:
The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .
Explanation:
V=40m/s
Vy=V.sina=40.sin20=40 . 0.342=13.68m/s
Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s
Projectile travels during 5 seconds and the ramge becomes:
x=V.t=30.64 . 5=153.2m
