A sled with mass 15 kg moves in a straight line on a horizontal surface as it is being pulled by a force F at an angle of 30° with the direction of motion. A force of friction of 10 N acts on the sled. At one point in its path, its speed is 4.0 m/s; after it has traveled a distance 5.0 m beyond this point, its speed is 7.0 m/s.Use the work-energy theorem to find the total work done on the sled?
Find the pulling force F on the sled.

680 Kg.m/s

Explanation:

Mass of player; m_p = 105 kg.

Speed of player before Collision; v_pi = 8.5 m/s

Mass of referee; m_r = 85 kg

Speed of referee before collision; v_ri = 3.5 m/s

Speed of referee after collision; v_rf = 6 m/s

From conservation of momentum,

Initial momentum = final momentum

Thus;

(m_p × v_pi) + (m_r × v_ri) = (m_p × v_pf) + (m_r × v_rf)

Where (m_p × v_pf) is the momentum of the player after collision.

Thus, Plugging in the relevant values, we have;

(105 × 8.5) + (85 × 3.5) = (m_p × v_pf) + (85 × 6)

(m_p × v_pf) = (105 × 8.5) + (85 × 3.5) - (85 × 6)

(m_p × v_pf) = 680 Kg.m/s