A child pulls a 3kg sled 200m along the ice by exerting a force of 40 N on a rope that makes an angle of
30 degrees with the horizontal. The coefficient of friction between the sled and the ice is 0.15.
a) How much work has the applied force done on the sled?
b) How much work has friction done on the sled?
c) What is the net work done on the sled?
a) The work done by the applied force can be calculated using the formula:
Work = Force × Distance × cos(angle)
Plugging in the values:
Work = 40 N × 200 m × cos(30°)
= 40 N × 200 m × (√3/2)
= 40 N × 200 m × 0.866
= 6928.2 J
Therefore, the work done by the applied force on the sled is 6928.2 Joules.
b) The work done by friction can be calculated using the formula:
Work = Force of friction × Distance
The force of friction can be calculated using the formula:
Force of friction = coefficient of friction × Normal force
The Normal force can be calculated as the weight of the sled, which is:
Normal force = mass × gravitational acceleration
= 3 kg × 9.8 m/s^2
= 29.4 N
Therefore, the force of friction is:
Force of friction = 0.15 × 29.4 N
= 4.41 N
Plugging in the values:
Work = 4.41 N × 200 m
= 882 J
Therefore, the work done by friction on the sled is 882 Joules.
c) The net work done on the sled can be calculated by subtracting the work done by friction from the work done by the applied force:
Net work = Work applied force - Work friction
= 6928.2 J - 882 J
= 6046.2 J
Therefore, the net work done on the sled is 6046.2 Joules.