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Amy is in-line skating. Her mass is 50 kg. She is rolling forward (north) on a flat section of road at 10 m/s. The rolling friction acting on the wheels of her skates is 10 N backward (south). Air resistance creates a 15 N force also acting backward (south) on her. If these are the only horizontal forces acting on Amy. What is her horizontal acceleration as a result of these forces?

Sagot :

Cricetusidn

Answer:

The right solution is "0.50 m/s²". A further explanation is provided below.

Explanation:

The given values are:

Mass,

m = 50 kg

Speed,

v = 10 m/s

Rolling friction acting backward (south),

f = 10 N

Air resistance acting backward (south),

[tex]F_T[/tex] = 15 N

The total force acting will be:

⇒  [tex]F = -f-F[/tex]

On substituting the given values, we get

⇒      [tex]=-10-15[/tex]

⇒      [tex]=-25 \ N[/tex]

Now,

⇒  [tex]a = \frac{F}{m}[/tex]

On substituting the given values, we get

⇒     [tex]=\frac{-25}{50}[/tex]

⇒     [tex]=-0.50 \ m/s^2[/tex]

The horizontal acceleration will be "0.50 m/s²" because the (-)ve sign indicates it in south direction.

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