Question

A teenager of mass m1 = 72 kg pushes backward against the ground with his foot as he rides his skateboard. This exerts a horizontal force of magnitude Ffoot = 11 N. The skateboard has m2 = 2.8 kg.

Answer 1

Answer:

The acceleration of the skateboard is -9.65 m/s².

Explanation:

Given that,

Mass of teenager = 72 kg

Force $F_{foot}=11\ N$

Mass of skateboard = 2.8 kg

Suppose, the determine the numerical value for the magnitude of the acceleration of skateboard

We need to calculate the acceleration of skateboard

Using balance equation

$F-(m_{1}+m_{2})g=(m_{1}+m_{2})a$

Put the value into the formula

$11-(72+2.8)\times9.8=(72+2.8)\times a$

$a=\dfrac{11-(72+2.8)\times9.8}{(72+2.8)}$

$a=-9.65\ m/s^2$

Hence, The acceleration of the skateboard is -9.65 m/s².

Answer 2

The acceleration of teenager with skateboard is $9.65m/s^{2}$

Acceleration:

It is defined as rate of change of velocity with respect to time.

Given that, mass of teenager $m_{1}=72kg$ and mass of skateboard $m_{2}=2.8kg$

• We have to find acceleration.

• The value of gravitational acceleration $g=9.8m/s^{2}$

• The balanced equation of force is,

                  $F-(m_{1}+m_{2})g=(m_{1}+m_{2})a$

• Substitute values in above expression.

                  $11-(72+2.8)*9.8=(72+2.8)*a\\\\a=-\frac{722.04}{74.8} =-9.65m/s^{2}$

The acceleration of teenager with skateboard is $9.65m/s^{2}$

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