A) The work done by the boy is 22.0 J.
B) The work done by friction is 15.07 J.
C) The final speed of the sled is 2.67 m/s.
D) The power developed by the boy is 11.0 W.
A) Work done by the boy
The work done by the boy is the product of the force he applies and the distance the sled moves in the direction of the force.
The horizontal component of the force is the only component responsible for the displacement of the sled.
Calculate the horizontal component of the force:
Force_horizontal = Force_applied * cos(theta)
= 11.0 N * cos(0°) // Assuming the force is applied horizontally
= 11.0 N
Calculate the work done by the boy:
Work_boy = Force_horizontal * distance
= 11.0 N * 2.0 m
= 22.0 J
B) Work done by friction
The work done by friction is the product of the force of friction and the distance the sled moves. The force of friction acts in the opposite direction to the displacement of the sled.
Calculate the normal force:
Normal_force = mass_sled * gravity
= 6.40 kg * 9.81 m/s²
= 62.784 N
Calculate the force of friction:
Friction_force = coefficient_friction * Normal_force
= 0.12 * 62.784 N
= 7.534 N
Calculate the work done by friction:
Work_friction = Friction_force * distance
= 7.534 N * 2.0 m
= 15.07 J
C) Final speed of the sled
The final speed of the sled can be calculated using the work-energy theorem:
Work_net = ΔKE
Work_boy + Work_friction = 1/2 * mass_sled * (v_final^2 - v_initial^2)
22.0 J - 15.07 J = 1/2 * 6.40 kg * (v_final^2 - 0.5 m/s)^2
6.93 J = 3.2 kg * (v_final^2 - 0.25 m/s^2)
v_final^2 = 7.185 m^2/s^2
v_final = √7.185 m^2/s^2
v_final = 2.67 m/s
D) Power developed by the boy
Power is the rate at which work is done.
Power = Work / ti
= 22.0 J / 2.0 s // Assuming the boy exerts the force for 2 seconds
= 11.0 W