A 2.6 kg block of ice at a temperature of 0.0�C

and an initial speed of 5.3 m/s slides across a
level floor.
If 3.3 × 105 J are required to melt 1.0 kg
of ice, how much ice melts, assuming that the
initial kinetic energy of the ice block is entirely
converted to the ice’s internal energy?

KE=1/2mv^2

KE=1/2(2.5kg)(5.7m/s)
KE=40.6J=1.230x10^-4kg

To solve this problem, we need to calculate the kinetic energy of the ice block and then compare it to the energy required to melt a certain amount of ice. Here are the steps to find out how much ice melts:

Step 1: Calculate the initial kinetic energy of the ice block.
The formula for kinetic energy is given by KE = (1/2) * m * v^2, where m is the mass and v is the velocity.

Given:
Mass of the ice block (m) = 2.6 kg
Initial velocity (v) = 5.3 m/s

Substituting the values into the formula:
KE = (1/2) * 2.6 kg * (5.3 m/s)^2
KE = 36.77 J

So, the initial kinetic energy of the ice block is 36.77 J.

Step 2: Calculate the amount of ice that can be melted.
Given:
Energy required to melt 1 kg of ice = 3.3 × 10^5 J

To find the amount of ice melted, we can divide the initial kinetic energy of the ice block by the energy required to melt 1 kg of ice.

Amount of ice melted = (Initial kinetic energy of the ice block) / (Energy required to melt 1 kg of ice)
Amount of ice melted = 36.77 J / (3.3 × 10^5 J/kg)

Calculating the value:
Amount of ice melted = 0.000111 kg

Therefore, assuming that the initial kinetic energy of the ice block is entirely converted to the ice's internal energy, approximately 0.000111 kg of ice will melt.