A 30 kg table is pulled by a constant force across a rough horizontal floor with an estimated coefficient of kinetic friction (μk) of 0.7. How much work is done by the friction force on the floor after being pulled for 30 m?
The work done by the friction force is given by:
Wfrict = μk * N * d
where μk is the coefficient of kinetic friction, N is the normal force (equal to the weight of the table, which is W = m*g), and d is the distance moved.
We first calculate the weight of the table:
W = m*g
W = 30 kg * 9.81 m/s^2
W = 294.3 N
The normal force is equal in magnitude and opposite in direction to the weight, so:
N = -W
N = -294.3 N
The negative sign indicates that the normal force is directed upwards, while the weight is directed downwards.
Now we can calculate the work done by the friction force:
Wfrict = μk * N * d
Wfrict = 0.7 * (-294.3 N) * 30 m
Wfrict = -6198.9 J
The negative sign indicates that the friction force is acting in the opposite direction to the displacement of the table, i.e. it is opposing the motion. The work done by the friction force is therefore negative, indicating that it is removing energy from the system (in the form of heat and sound).