Question

If I am pulling a bag with a force of 25 N and the bag is on an angle of 32 degrees from the ground and the bag has a 2 kg weight, what is the coefficient of kinetic friction between the bag and the ground?

Answer 1

If I am pulling a bag with a force of $25\mathrm{N}$ and the bag is on an angle of 32 degrees from the ground and the bag has a $2\mathrm{kg}$ weight, what is the coefficient of kinetic friction between the bag and the ground?

$\alpha = 32$

$mg = 9.8\frac{m}{s^2} 2kg = 19.6H$

$F = 25H$ - force of pulling

$F_{r}$ force of friction

From Newton's first law of motion:

If there is no net force on an object, then its velocity is constant.

So, vector sum of forces equals 0.

$\mu (or k) = F_{f} / N$ - the coefficient of kinetic friction between the bag and the ground

Therefore:

$F_{f} + mg\sin \alpha = F \Rightarrow F_{f} = F - mg\sin \alpha$

and:

$mg\cos \alpha = N \Rightarrow N = mg\cos \alpha$

$\mu (k) = \frac{F_f}{N} = \frac{F - mg\sin\alpha}{mg\cos\alpha} = \frac{25 - 19.6\sin 32}{19.6\cos 32} = 0.88$

Answer: the coefficient of kinetic friction between the bag and the ground equals 0.88