Respuesta :
Answer:
3.97 m/s
Explanation:
given,
mass of the object,m= 2 Kg
angle of inclination,θ = 20°
coefficient of friction,μ= 0.2
distance,L = 4 m
using work energy theorem
[tex]W = \dfrac{1}{2}mv^2[/tex]..........(1)
now,
v is the speed of the box.
W is the work done on the box by a net external force.
Work done on the force
[tex]W = F_{net}L[/tex]
[tex]F_{net} = m g sin \theta - \mu N[/tex]
[tex]F_{net} = m g sin \theta - \mu (mg cos \theta)[/tex]
[tex]W = (mg sin\theta - \mu mg cos \theta)L[/tex]......(2)
now, equating both equation 1 and 2
[tex]\dfrac{1}{2}mv^2 = (mg sin\theta - \mu mg cos \theta)L[/tex]
[tex]v= \sqrt{2 (g sin\theta - \mu g cos \theta)L}[/tex]
[tex]v= \sqrt{2(9.8\times sin 20^0 - 0.15\times 9.8 \times cos 20^0)4}[/tex]
v = 3.97 m/s
hence, the final speed of the box is equal to v = 3.97 m/s
The final speed the mass after sliding 4.0 meters along the plane is 3.48 m/s.
Given the following data:
- Mass = 2.0 kg.
- Angle = 20°
- Coefficient of kinetic friction = 0.20.
- Distance = 4.0 m.
Scientific data:
- Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
To calculate the final speed the mass after sliding 4.0 meters along the plane, we would apply the work-energy theorem:
How to calculate the work done.
Mathematically, the work done on the block of ice is given by this formula:
[tex]Work =Fd[/tex] .....equation 1.
But, [tex]F = mgsin \theta - \mu mgcos \theta[/tex]
So, the work done becomes;
[tex]W = (mgsin \theta - \mu mgcos \theta)d[/tex] ....equation 2.
From the work-energy theorem, the work done is given by this formula:
[tex]W =\frac{1}{2} mv^2[/tex] ....equation 3.
Equating eqn. 3 and eqn. 2, we have:
[tex]\frac{1}{2} mv^2 = (mgsin \theta - \mu mgcos \theta)d\\\\V=\sqrt{2(gsin \theta - \mu gcos \theta)d} \\\\V=\sqrt{2(9.8 \times sin 20- 0.20 \times 9.8 \times cos 20)\times 4.0}\\\\V=\sqrt{2(3.3518- 1.8418)\times 4.0}\\\\V=\sqrt{2(1.51)\times 4.0}\\\\V=\sqrt{12.08}[/tex]
V = 3.48 m/s.
Read more on work here: https://brainly.com/question/22599382
