Question

A crate having mass 50.0 kg falls horizontally off the back of the flatbed truck, which is traveling at 100 km/h. Find the value of the coefficient of kinetic friction between the road and crate if the crate slides 50 m on the road in coming to rest. The initial speed of the crate is the same as the truck, 100 km/h.

(a) 0.30
(b) 0.25
(c) 0.35
(d) 0.20

Answer 1

(c) 0.35

Option "c" determined by equating the work done by kinetic friction to the initial kinetic energy and solving for the coefficient of kinetic friction, resulting in a value of approximately 0.35.

Step-by-step explanation:

The coefficient of kinetic friction can be determined using the work-energy principle. When the crate comes to rest, the work done by friction is equal to the initial kinetic energy of the crate.

The work done by friction can be expressed as the product of the frictional force, the displacement, and the cosine of the angle between the force and displacement vectors.

`\[ W_{\text{friction}} = \text{Frictional Force} * \text{Displacement} * \cos(\theta) \]`

In this case, the crate moves horizontally, so the angle
`(\(\theta\)) \\`between the frictional force and displacement is 0 degrees, and
`\(\cos(0^\circ) = 1\).`Thus, the equation simplifies to:

`\[ W_{\text{friction}} = \text{Frictional Force} * \text{Displacement} \]`

The work done by friction is also equal to the initial kinetic energy of the crate:

`\[ W_{\text{friction}} = (1)/(2) m v_{\text{initial}}^2 \]`

Now, we can set these two expressions for work equal to each other:

`\[ \text{Frictional Force} * \text{Displacement} = (1)/(2) m v_{\text{initial}}^2 \]`

Solving for the frictional force and dividing by the weight of the crate gives the coefficient of kinetic friction:

`\[ \text{Coefficient of Kinetic Friction} = \frac{\text{Frictional Force}}{\text{Weight}} \]`

Substituting the known values into the equation, we find that the coefficient of kinetic friction is approximately \(0.35\), matching option (c).