Answer :
To find the force of kinetic friction, we start by calculating the normal force that Gary experiences due to his weight. The normal force for an object on a horizontal surface is given by
[tex]$$
F_N = m \times g,
$$[/tex]
where:
- [tex]$m$[/tex] is the mass of the object,
- [tex]$g$[/tex] is the acceleration due to gravity.
Given that Gary's mass is [tex]$52\ \text{kg}$[/tex] and using [tex]$g = 9.8\ \text{m/s}^2$[/tex], we have
[tex]$$
F_N = 52\ \text{kg} \times 9.8\ \text{m/s}^2 = 509.6\ \text{N}.
$$[/tex]
Next, the force of kinetic friction ([tex]$F_k$[/tex]) is calculated using the coefficient of kinetic friction ([tex]$\mu_k$[/tex]) and the normal force ([tex]$F_N$[/tex]):
[tex]$$
F_k = \mu_k \times F_N.
$$[/tex]
Given that [tex]$\mu_k = 0.28$[/tex], the friction force is
[tex]$$
F_k = 0.28 \times 509.6\ \text{N} \approx 142.688\ \text{N}.
$$[/tex]
Rounding to the nearest whole number, the force of kinetic friction is approximately [tex]$143\ \text{N}$[/tex].
Thus, the correct answer is:
A. [tex]$143\ \text{N}$[/tex].
[tex]$$
F_N = m \times g,
$$[/tex]
where:
- [tex]$m$[/tex] is the mass of the object,
- [tex]$g$[/tex] is the acceleration due to gravity.
Given that Gary's mass is [tex]$52\ \text{kg}$[/tex] and using [tex]$g = 9.8\ \text{m/s}^2$[/tex], we have
[tex]$$
F_N = 52\ \text{kg} \times 9.8\ \text{m/s}^2 = 509.6\ \text{N}.
$$[/tex]
Next, the force of kinetic friction ([tex]$F_k$[/tex]) is calculated using the coefficient of kinetic friction ([tex]$\mu_k$[/tex]) and the normal force ([tex]$F_N$[/tex]):
[tex]$$
F_k = \mu_k \times F_N.
$$[/tex]
Given that [tex]$\mu_k = 0.28$[/tex], the friction force is
[tex]$$
F_k = 0.28 \times 509.6\ \text{N} \approx 142.688\ \text{N}.
$$[/tex]
Rounding to the nearest whole number, the force of kinetic friction is approximately [tex]$143\ \text{N}$[/tex].
Thus, the correct answer is:
A. [tex]$143\ \text{N}$[/tex].
