It has no units. <span>
The coefficient of friction is dimensionless:
[force]/[force] = Newton/Newton = 1</span>
Send a pic or something so more people understand
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
1) 29.5 N/m
2) 0.100 m
Explanation:
1)
The force constant of the spring can be found by using the fact that the force on the spring is proportional to the extension of the spring (Hooke's Law). Therefore, we can write:
where
is the change in the force on the spring, where
is the force applied when the hanging mass is
is the force applied when the hanging mass is
is the change in extension of the spring, where
is the extension of the spring when the hanging mass is 0.300 kg
is the extension of the spring when the hanging mass is 1.95 kg
Solving for k,
2)
When the first mass is hanging on the spring, we have
where:
is the force applied on the spring (the weight of the hanging mass)
k is the spring constant
is the extension of the spring wrt its natural length
is the natural length of the spring (the unloaded length)
Here we have
k = 29.5 N/m
Solving for , we find:
