A box is sliding on a rough patch of ice, and the only horizontal force acting on it is friction. You observe that it slides to a stop in 600 seconds over a distance of 23 meters. What magnitude coefficient of kinetic friction between the box and the ice?

The magnitude of the coefficient of kinetic friction is approximately 1.304 × 10⁻⁵

Explanation:

The given parameters are;

The time it takes for the box to stop, "t" = 600 seconds

The distance the box moves before coming to a stop, "d" = 23 meters

The work done by the friction force to stop the block = The kinetic energy lost by the box

The work done by the friction force to stop the block = ·m·g·d

The kinetic energy change of the block = 1/2·m·u²

Where;

g = The acceleration due to gravity = 9.8 m/s²

m = The mass of the box

= The coefficient of kinetic friction

d = The distance travelled by the block = 600 m

u = The initial velocity of the block

∴ ·m·g·d = 1/2·m·u²

·g·d = 1/2·u²

= 1/2·u²/(g·d) = 1/2·u²/(9.8 × 23) = u²/450.8

∴   = u²/450.8...(1)

The kinematic equation of motion of the block can be presented as follows;

v = u - a·t

Where;

v = The final velocity = 0 m/s (The block slides to a stop)

u = The initial velocity

a = The acceleration of the block (the minus sign indicates that the block's velocity is decreasing)

t = The time of motion of the block = 600 seconds

Therefore, by substituting the above values, we have;

0 = u - 600·a

∴ u = a·600

The kinematic equation of motion for the distance moved by the block, "s" can be expressed as follows;

s = u·t - 1/2·a·t²

Where;

s = The distance moved by the bloc = 23 meters

∴ 23 = a·600·600 - 1/2·a·600² = 360,000·a - 1/2·a·360,000 = 180,000·a

a = 23/(180,000)

u = 600 × (23/(180,000)) = 23/300

The initial velocity, u = 23/300 m/s

From equation (1), we have;

= u²/(450.8) = (23/300)²/450.8 = 23/1764000 ≈ 1.304 × 10⁻⁵

The magnitude of the coefficient of kinetic friction, ≈ 1.304 × 10⁻⁵

ice to water 1 water to vapour 2 latent heat