Answer :
Answer:
66.944 Joules
66.944 Joules
0.457 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
k = Spring constant = 640 N/m
m = Mass of block = 3.5 kg
x = Compression
[tex]v^2-u^2=2as\\\Rightarrow -u^2=2\mu gs-v^2\\\Rightarrow u=\sqrt{v^2-2\mu gs}\\\Rightarrow u=\sqrt{0^2-2\times 0.25\times -9.81\times 7.8}\\\Rightarrow u=6.185\ m/s[/tex]
[tex]Q=\frac{1}{2}m(v^2-u^2)\\\Rightarrow Q=\frac{1}{2}3.5(0^2-6.185^2)\\\Rightarrow Q=-66.944\ Joules[/tex]
Increase in thermal energy is 66.944 Joules
Kinetic energy of the block is 66.944 Joules
[tex]\frac{1}{2}m(v^2-u^2)=\frac{1}{2}kx^2\\\Rightarrow x=\sqrt{\frac{66.944\times 2}{640}}\\\Rightarrow x=0.457\ m[/tex]
Original compression of the spring is 0.457 m
The thermal energy of the object is the sum of its potential and kinetic energy.
A) The increase in the thermal energy of the system is 66.95 Joules.
B) The kinetic energy of the block is 66.95 Joules.
C) The original compression distance of the spring is 0.45 m.
How do you calculate the thermal energy?
Given that mass of the block is 3.5 kg, the compression constant of the spring is 640 N/m. The kinetic friction coefficient is 0.25. The final velocity of the block is zero after a distance 7.8 m.
Part A.
The velocity can be calculated as given below.
[tex]v^2 = u^2 +2as[/tex]
[tex]v^2=u^2+2kgs[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration, g is the gravitational acceleration, k is the compression constant and s is the distance.
[tex]0^2 = u^2 + 2\times 0.25\times -9.8\times 7.8[/tex]
[tex]u = \sqrt{38.22}[/tex]
[tex]u=6.185 \;\rm m/s[/tex]
The block is accelerated hence its total thermal energy is equal to its kinetic energy.
[tex]TE = \dfrac{1}{2} m (v^2-u^2)[/tex]
[tex]TE = \dfrac {1}{2}\times 3.5\times (0 - 6.185^2)[/tex]
[tex]TE = - 66.95 \;\rm J[/tex]
Hence we can conclude that the increase in thermal energy is 66.95 Joules.
Part B.
The kinetic energy of the block is 66.95 Joules.
Part C.
The compression distance of the spring can be calculated as given below.
[tex]\dfrac {1}{2}\times m\times(v^2-u^2) = \dfrac {1}{2}\times k\times d^2[/tex]
[tex]\dfrac {1}{2}\times 3.5\times (0-6.185^2) = \dfrac {1}{2}\times 640\times d^2[/tex]
[tex]d=0.45 \;\rm m[/tex]
Hence we can conclude that the original compression distance of the spring is 0.45 m.
To know more about thermal energy, follow the link given below.
https://brainly.com/question/11278589.