Answer:
6.05 m/s
Explanation:
In order for the puck to reach the top of the ramp, its initial kinetic energy must be equal to its final potential energy.
So we can write:
[tex]KE=PE\\\frac{1}{2}mv^2 = mgh[/tex] (1)
where
m = 100 g = 0.1 kg is the mass of the puck
v is the initial speed of the puck
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
h is the height of the ramp
Here we know that
d = 4.60 m is the length of the ramp
[tex]\theta=24.0^{\circ}[/tex] is the angle of the ramp
So its height is
[tex]h=d sin \theta = (4.60)(sin 24.0^{\circ})=1.87 m[/tex]
So now we can re-arrange eq (1) to find the minimum speed of the puck:
[tex]v=\sqrt{2gh}=\sqrt{2(9.8)(1.87)}=6.05 m/s[/tex]
The minimum speed that a 100 g puck needs to make it to the top of a 4.60 m -long, 24.0 ∘ frictionless ramp is 6.05m/s
The kinetic energy along the ramp is equal to the potential energy at the top.
KE = PE
[tex]\frac{1}{2}mv^2=mgh\\ \frac{1}{2}v^2 = gh\\v^2=2gh\\v =\sqrt{2gh}[/tex]
Given the following parameters
g is the acceleration due to gravity = 9.8m/s²
h = 4.60sin24⁰
h = 1.870m
Get the minimum speed required
[tex]v =\sqrt{2(9.8)(1.87)} \\v =\sqrt{36.652}\\v= 6.05m/s[/tex]
Hence the minimum speed that a 100 g puck needs to make it to the top of a 4.60 m -long, 24.0 ∘ frictionless ramp is 6.05m/s
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