Respuesta :
Given :
A sled and its rider are moving at a speed of 2.2 m/s along a horizontal stretch of snow.
The coefficient of kinetic friction is 0.050.
To Find :
The displacement of the sled.
Solution :
Let, total mass of sled and rider is m.
So, frictional force is :
[tex]F = \mu mg[/tex]
Let, the acceleration of body is a.
So,
[tex]\mu mg = ma\\\\a = \mu g\\\\a=0.050\times 10\\\\a= 0.5\ m/s^2[/tex]
We know, by equation of motion :
[tex]2as=v^2-u^2\\\\s = \dfrac{0-2.2^2}{2\times 0.5}\\\\s=-4.84\ m[/tex]
Therefore, displacement of sled is -4.84 m.
Hence, this is the required solution.
The displacement of the sled is -4.84m
Equation of motion:
Since the speed is 2.2 m/s with a horizontal stretch of snow. And, there is The coefficient of kinetic friction i.e. 0.050.
So, here we assume the total mass of sled and rider be m
And, the acceleration should be a
So,
umg = ma
a = ug
a = 0.50*10
= 0.5m/s^2
Here we applied the equation of motion
So, 2as = v^2 - u^2
s = (0-2.2^2) / (2*0.5)
= -4.84m
Therefore, The displacement of the sled is -4.84m
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