you draw a 100kg sack of maize meal across the floor of a warehouse for the distance of 10m. You apply a force of 300N, directed at an angle of 30 degree to the floor.
a) If the floor is frictionless, how much work do you do on the bag?
b) If the coefficient of friction between the sack and the bag is 0.20, how much work is done on the sack?
c) If the sack starts from rest, what its kinetic energy at the end of the 10m in case b?
To answer these questions, we need to understand the concept of work and how it relates to force, distance, and energy.
a) If the floor is frictionless, the work done on the sack can be calculated using the formula:
Work = Force * Distance * cos(theta)
where Force is the magnitude of the applied force, Distance is the displacement of the sack, and theta is the angle between the force vector and the direction of displacement.
In this case, Force = 300N, Distance = 10m, and theta = 30 degrees.
Calculating the work done:
Work = 300N * 10m * cos(30 degrees)
= 300N * 10m * (√3/2)
= 4500J (Joules)
Therefore, the work done on the sack is 4500 Joules.
b) If there is a coefficient of friction between the sack and the bag, the work done on the sack can be calculated using a modified formula:
Work = Force * Distance * cos(theta) - Frictional Force * Distance
In this case, we need to calculate the frictional force.
Frictional Force = Coefficient of Friction * Normal Force
The Normal Force is equal to the weight of the sack, which is given as 100kg * 9.8 m/s^2 (acceleration due to gravity) = 980N.
Plugging in the values:
Frictional Force = 0.20 * 980N = 196N
Now, calculating the work done:
Work = (300N * 10m * cos(30 degrees)) - (196N * 10m)
= 300N * 10m * (√3/2) - 196N * 10m
= 2530J (Joules)
Therefore, the work done on the sack with friction is 2530 Joules.
c) If the sack starts from rest and we want to find its kinetic energy at the end of the 10m, we need to understand that work done on an object is equal to the change in its kinetic energy.
Therefore, the work done on the sack with friction (2530 J) is equal to the change in its kinetic energy.
Initial Kinetic Energy = 0 (as it starts from rest)
Change in Kinetic Energy = Work Done on the sack with friction = 2530 J
Final Kinetic Energy = Initial Kinetic Energy + Change in Kinetic Energy
Substituting the values:
Final Kinetic Energy = 0 + 2530 J
= 2530 J
Therefore, the kinetic energy of the sack at the end of the 10m distance in case b is 2530 Joules.