Question 83: Laws of Motion

Question

A heavy box of mass $50 \mathrm{~kg}$ is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is 0.3 then force of kinetic friction is :

Accepted Answer

To find the force of kinetic friction acting on the box, we can use the formula for kinetic friction, which is given by:
$F_{	ext{friction}} = \mu_{k} \cdot N$
where:
$F_{	ext{friction}}$ is the force of friction,
$\mu_{k}$ is the coefficient of kinetic friction,
$N$ is the normal force exerted by the surface onto the box.
Since the box is moving on a horizontal surface, the normal force $N$ would be equal to the weight of the box, which is calculated by $mg$, where $m$ is the mass of the box and $g$ is the acceleration due to gravity. For most calculations, $g$ is approximated as $9.8 \, 	ext{m/s}^2$.
Substituting the given values:
$m = 50 \, 	ext{kg}$,
$\mu_k = 0.3$,
$g = 9.8 \, 	ext{m/s}^2$.
First, calculate the weight of the box, which is the normal force:
$N = mg = 50 \, 	ext{kg} 	imes 9.8 \, 	ext{m/s}^2 = 490 \, 	ext{N}$
Then, use this value to find the force of kinetic friction:
$F_{	ext{friction}} = \mu_{k} \cdot N = 0.3 	imes 490 \, 	ext{N} = 147 \, 	ext{N}$
Thus, the force of kinetic friction acting on the box is
147 N
. The correct answer is
Option B
.

Answer

1470 N

Answer

147 N

Answer

1.47 N

Answer

14.7 N

Laws of Motion

A heavy box of mass 50 \mathrm{~kg} is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is 0.3 then force of kinetic friction is :

Choose the correct answer: