Answer:
![x=0.0049\ m= 4.9\ mm](https://tex.z-dn.net/?f=x%3D0.0049%5C%20m%3D%204.9%5C%20mm)
![d=0.01153\ m=11.53\ mm](https://tex.z-dn.net/?f=d%3D0.01153%5C%20m%3D11.53%5C%20mm)
Explanation:
Given:
- mass of the object,
![m=0.75\ kg](https://tex.z-dn.net/?f=m%3D0.75%5C%20kg)
- elastic constant of the connected spring,
![k=150\ N.m^{-1}](https://tex.z-dn.net/?f=k%3D150%5C%20N.m%5E%7B-1%7D)
- coefficient of static friction between the object and the surface,
![\mu_s=0.1](https://tex.z-dn.net/?f=%5Cmu_s%3D0.1)
(a)
Let x be the maximum distance of stretch without moving the mass.
<em>The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.</em>
![f=k.x](https://tex.z-dn.net/?f=f%3Dk.x)
![\mu_s.N=k.x](https://tex.z-dn.net/?f=%5Cmu_s.N%3Dk.x)
where:
N = m.g = the normal reaction force acting on the body under steady state.
![0.1\times (9.8\times 0.75)=150\times x](https://tex.z-dn.net/?f=0.1%5Ctimes%20%289.8%5Ctimes%200.75%29%3D150%5Ctimes%20x)
![x=0.0049\ m= 4.9\ mm](https://tex.z-dn.net/?f=x%3D0.0049%5C%20m%3D%204.9%5C%20mm)
(b)
Now, according to the question:
- Amplitude of oscillation,
![A= 0.0098\ m](https://tex.z-dn.net/?f=A%3D%200.0098%5C%20m)
- coefficient of kinetic friction between the object and the surface,
![\mu_k=0.085](https://tex.z-dn.net/?f=%5Cmu_k%3D0.085)
Let d be the total distance the object travels before stopping.
<em>Now, the energy stored in the spring due to vibration of amplitude:</em>
![U=\frac{1}{2} k.A^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7D%20k.A%5E2)
<u><em>This energy will be equal to the work done by the kinetic friction to stop it.</em></u>
![U=F_k.d](https://tex.z-dn.net/?f=U%3DF_k.d)
![\frac{1}{2} k.A^2=\mu_k.N.d](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20k.A%5E2%3D%5Cmu_k.N.d)
![0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d](https://tex.z-dn.net/?f=0.5%5Ctimes%20150%5Ctimes%200.0098%5E2%3D0.0850%20%5Ctimes%200.75%5Ctimes%209.8%5Ctimes%20d)
![d=0.01153\ m=11.53\ mm](https://tex.z-dn.net/?f=d%3D0.01153%5C%20m%3D11.53%5C%20mm)
<em>is the total distance does it travel before stopping.</em>
Answer:
The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.
Explanation:
Answer:The velocity of the object will be 5
m/s or 13.23m/s
Explanation:
force exerted by the object= 30N
distance displayed by the object by the action of force=6.0m
mass of object=10kg
velocity gained by the object=?
![\frac{1}{2}mv^{2}= forcexdisplacement\\\frac{1}{2}10v^{2} = 30x6\\ 5v^{2}=180\\ v^{2}= 180-5\\ v^{2} =175\\v=\sqrt{175} \\v=5\sqrt{7} or 13.23](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D%3D%20forcexdisplacement%5C%5C%5Cfrac%7B1%7D%7B2%7D10v%5E%7B2%7D%20%3D%2030x6%5C%5C%205v%5E%7B2%7D%3D180%5C%5C%20v%5E%7B2%7D%3D%20180-5%5C%5C%20v%5E%7B2%7D%20%3D175%5C%5Cv%3D%5Csqrt%7B175%7D%20%5C%5Cv%3D5%5Csqrt%7B7%7D%20or%2013.23)
Answer:
Explanation:
Given that,
Current measure is
i=10±0.6 Amps
And also,
R=45.0±2.0 Ω
Power dissipated by
P=i²R
Then
P=(10±0.6)²(45.0±2.0)
P=10²×45
P=450Watts
Now, calculating the uncertainty
∆P=|P| • √(2(∆i/i)²+(∆R/R)²)
∆P=450√ (2×(0.6/10)²+(2/45)²)
∆P=450√(0.0072+0.001975)
∆P=450√0.009175
∆P=43.1
The uncertainty in power is 43.1
Then,
P=450 ± 43.1 Watts
Answer:
0.15m i tink blablablablabla