Respuesta :
Answer:
10.71m/s
Explanation:
Mechanical energy is conserved since the ramp surface had no friction and the experiences no significant air resistance
Potential energy at the top of the ramp = mgh
where m is m, g = 9.8 m/s² ( acceleration due to gravity ) and h is height
since the length of the surface L of the ramp is 20.0 m and it is inclined at 17°
h, height = Lsinθ = 20 sin 17° = 5.85 m
kinetic energy of the block = [tex]\frac{1}{2}[/tex]mv²
potential energy = kinetic energy
m × 9.8 × 5.85 m = [tex]\frac{1}{2}[/tex]mv²
cancel m on both side
v = √(2 × 9.8 × 5.85) = 10.71m/s
The speed of the block as it reaches the bottom of the ramp will be 10.71m/s.
What is speed ?
Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.
The given data in the problem is;
[tex]\rm \theta[/tex] is the ramp angle = 17⁰.
L is the length of the surface of the ramp i=20.0 m
v is the speed of the block =?
Because the ramp surface has little friction and there is no considerable air resistance, mechanical energy is saved.At the top of the ramp, there is a lot of potential energy.
PE = mgh
From the trignometry the length is found as;
h= Lsinθ
h= 20 sin 17°
h= 5.85 m
The kinetic energy of the block is found as;
[tex]\rm KE = \frac{1}{2} mv^2[/tex]
From the law of conservation of energy the kinetic energy is equal to the potential energy;
[tex]\rm mgh= \frac{1}{2} mv^2[/tex]
[tex]\rm v= \sqrt{2gh} \\\\ \rm v= \sqrt{2\times 9.81 \times 5.85 } \\\\ \rm v= 10.71 \ m/sec[/tex]
Hence the speed of the block as it reaches the bottom of the ramp will be 10.71m/s.
To learn more about the speed refer to the link;
https://brainly.com/question/7359669
