For the girl pulling a sled with a mass of 6.0 kg t the left with a force of 10.0 N at a 30.0° angle, we have:
a) The normal force acting on the sled is 294 N.
b) The acceleration of the sled is 1.2 m/s².
a) We can calculate the normal force acting on the sled, with the sum of all the forces acting in the vertical direction:
[tex] \Sigma F_{y} = 0 [/tex]
[tex] N + F_{y} - W = 0 [/tex]
[tex] N = W - F_{y} [/tex] (1)
Where:
W: is the weight force or force of gravity = 58.8 N
[tex]F_{y}[/tex]: is the component of the applied force in the y-direction
N: is the normal force =?
The vertical component of the force ([tex]F_{y}[/tex]) can be calculated as follows:
[tex] F_{y} = Fsin(\theta) [/tex] (2)
Where:
F: is the applied force = 10.0 N
θ: is the angle (with the horizontal) = 30.0°
By entering equation (2) into (1) we have:
[tex] N = W - Fsin(\theta) = 58.8 N*10.0N*sin(30) = 294 N [/tex]
Hence, the normal force acting on the sled is 294 N.
b) The acceleration of the sled can be found with the forces acting on the x-direction:
[tex] \Sigma F_{x} = ma [/tex]
[tex] F_{x} - F_{\mu} = ma [/tex]
Where:
a: is the acceleration =?
m: is the mass = 6.0 kg
[tex]F_{x}[/tex]: is the component of the applied force in the x-direction = Fcos(θ)
[tex]F_{\nu}[/tex]: is the force of friction = 1.5 N (it is negative because is in the opposite direction of motion)
Therefore, the acceleration is:
[tex] a = \frac{Fcos(\theta) - F_{\mu}}{m} = \frac{10.0 N*cos(30) - 1.5 N}{6.0 kg} = 1.2 m/s^{2} [/tex]
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