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Two forces are acting on a 0.250 kg hockey puck as it slides along the ice. The first force has a magnitude of 0.320 N and points 15.0° north of east. The second force has a magnitude of 0.600 N and points 55.0 ° north of east. If these are the only two forces acting on the puck, what will be the magnitude and direction of the puck's acceleration? Enter the direction as an angle measured in degrees counterclockwise from due east.

Respuesta :

Answer:

[tex]a = 3.45 m/s^2[/tex]

[tex]theta = 43.5 degree[/tex] North of East

Explanation:

two forces on the hockey puck is given as

[tex]F_1 = 0.320 N[/tex] at 15 degree North of East

[tex]F_1 = 0.320(cos15 \hat i + sin15\hat j)[/tex]

[tex]F_1 = 0.31 \hat i + 0.08 \hat j[/tex]

similarly other force is

[tex]F_2 = 0.60 N[/tex] at 55 degree North of East

[tex]F_2 = 0.60(cos55 \hat i + sin55 \hat j)[/tex]

[tex]F_2 = 0.34 \hat i + 0.49 \hat j[/tex]

Now net force on the puck is given as

[tex]F = F_1 + F_2[/tex]

[tex]F = (0.31 + 0.34)\hat i + (0.08 + 0.49)\hat j[/tex]

[tex]F = 0.65 \hat i + 0.57\hat j[/tex]

magnitude of force is given as

[tex]F = \sqrt{0.65^2 + 0.57^2}[/tex]

[tex]F = 0.86 N[/tex]

magnitude of acceleration

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{0.86}{0.25} [/tex]

[tex]a = 3.45 m/s^2[/tex]

direction of acceleration is given as

[tex]tan\theta = \frac{0.57}{0.60}[/tex]

[tex]theta = 43.5 degree[/tex] North of East

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