Question

Mass 26 kg is now sliding down a rough-surfaced incline that makes an angle of 27° with the horizontal. The coefficient of kinetic friction between the box and the surface is 0.36.

A) True
B) False
C) Cannot be determined
D) Depends on the surface type

Answer 1

The object will not accelerate down the incline, as the gravitational force component is not greater than the frictional force opposing the motion. Thus the correct answer is option B False.

Step-by-step explanation:

The object is sliding down a rough-surfaced incline, and we need to determine if the motion is constant. The key factor here is the angle of the incline and the coefficient of kinetic friction. The gravitational force component pulling the object down the incline is given by
`\(F_{\text{downhill}} = m \cdot g \cdot \sin(\theta)\),` where m is the mass,
`\(g\)`is the acceleration due to gravity, and
`\(\theta\)`is the angle of the incline.

The frictional force opposing the motion is
`\(F_{\text{friction}} = \mu_k \cdot N\),` where
`\(\mu_k\)` is the coefficient of kinetic friction, and \(N\) is the normal force. The normal force N is influenced by the weight of the object and the angle of the incline and is given by \(N = m \cdot g \cdot \cos(\theta)\).

If the gravitational force down the incline is greater than the frictional force, the object will accelerate. The condition for constant motion is when these two forces are equal. Therefore, if \(F_{\text{downhill}} > F_{\text{friction}}\), the object accelerates; if \(F_{\text{downhill}} = F_{\text{friction}}\), the object moves at a constant speed.

In this case, the frictional force
`\(F_{\text{friction}} = \mu_k \cdot N\),` and
`\(N = m \cdot g \cdot \cos(27^\circ)\). If \(\sin(27^\circ) > \mu_k \cdot \cos(27^\circ)\),`the object accelerates. Otherwise, it moves at a constant speed. Evaluating this condition, we find that
`\(\sin(27^\circ) > 0.36 \cdot \cos(27^\circ)\),`which is false. Therefore, the correct answer is B) False.