A block of mass 5 kg, which has an initial
speed of 3 m/s at time t = 0, slides on a
horizontal surface. Find the magnitude of the work that must be done on the block to bring it to rest.
Answer in units of J.
Negative work must be done ON the block, and the magnitude equals the initial kinetic energy, (M/2)*Vo^2
To find the magnitude of the work that must be done on the block to bring it to rest, we need to first calculate the kinetic energy of the block when it has come to rest.
The formula for kinetic energy is:
Kinetic energy (KE) = (1/2) * mass * velocity^2
Given:
mass (m) = 5 kg
initial velocity (v) = 3 m/s
We can calculate the initial kinetic energy (KEi) using the formula:
KEi = (1/2) * m * v^2
Substituting the given values:
KEi = (1/2) * 5 kg * (3 m/s)^2
= (1/2) * 5 kg * 9 m^2/s^2
= 45 J
Since the block comes to rest, all of its kinetic energy is converted into work. Therefore, the magnitude of the work done on the block to bring it to rest is equal to the initial kinetic energy (KEi):
Work = KEi
= 45 J
So, the magnitude of the work that must be done on the block to bring it to rest is 45 Joules (J).