Question

A teenager of mass m1 = 64 kg pushes backward against the ground with his foot as he rides his skateboard. This exerts a horizontal force of magnitude Ffoot = 13 N. The skateboard has m2 = 2.9 kg.

a. Write an expression for the magnitude of the horizontal component of force that the ground exerts on the teenager's foot, Fground.
b. Write an expression in terms of given quantities for the magnitude of the skateboard's acceleration, a, while the teenager is pushing backwards on the ground.
c. What is the numerical value for the magnitude of the acceleration, a, in m/s2?

Answer 1

We multiply the magnitude of the vector even by the cosine angle referenced towards the horizontal. The horizontal force defines as the force exerted inside a direction parallel to the horizon. Acceleration is merely the rate during which velocity changes. As a result, the magnitude indicates how rapidly velocity varies, and the further calculation can be defined as follows:

For option a:

$\to F_{foot}= 13 \ N\\\\\to m_1= 64 \ kg\\\\\to m_2=2.9 \ kg$

Please find the graph for the direction:

$\to |F_{ground}|= |F_{foot}| \\\\ \to F_{ground}= -F_{foot}= - 13 \ N$

For option b:  

$F_{ground}=(m_1+m_2)a\\\\a=\frac{F_{ground}}{m_1+m_2}$

For option c:  

$\bold{a=\frac{13}{64+2.9}}$

  $\bold{=\frac{13}{66.9}}\\\\\bold{=0.194 \ \frac{m}{s^2}}$

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