A horizontal force of magnitude 30.0 N pushes a block of mass 3.85 kg a distance of 3.00 m across a floor, where the coefficient of kinetic friction is 0.600
a.)How much work is done by that applied force on the block-floor system?
b.)During that displacement, the thermal energy of the block increases by 38.0 J. What is the increase in thermal energy of the floor?
c.)What is the increase in the kinetic energy of the block?
To find the answers to these questions, we'll need to use some formulas and equations related to work, friction, and energy.
a.) To find the work done by the applied force on the block-floor system, we'll use the formula:
Work = Force x Distance x Cosine(angle)
Since the force is applied horizontally, the angle between the force and the displacement is 0 degrees, so the cosine of the angle is 1.
Therefore, the work done by the applied force is:
Work = 30.0 N x 3.00 m x 1 = 90.0 J
b.) The increase in thermal energy of the block is given as 38.0 J. This increase is due to the work done against the frictional force between the block and the floor.
The work done against friction is given by the equation:
Work = Force of friction x Distance
The force of friction is given by:
Force of friction = coefficient of friction x Normal force
The normal force in this case is equal to the weight of the block:
Normal force = mass x gravity
The coefficient of kinetic friction is given as 0.600.
Therefore, we can calculate the force of friction as:
Force of friction = 0.600 x (3.85 kg x 9.8 m/s^2)
Now, using the calculated force of friction and the distance of 3.00 m, we can calculate the work done against friction:
Work = (0.600 x (3.85 kg x 9.8 m/s^2)) x 3.00 m
c.) The increase in kinetic energy of the block can be calculated using the work-energy theorem:
Increase in kinetic energy = Work done by the applied force
So, the increase in kinetic energy of the block is equal to the work done by the applied force, which we already calculated as 90.0 J.