Question

An 8.75 kg point mass and a 12.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point between the two masses 18.0 cm from the 8.75 kg mass along the line connecting the two fixed masses. For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Acceleration due to gravitational attraction. Find the magnitude a of the acceleration of the particle. Express your answer in meters per second squared. View Available Hint(s) I ASO ? a= m/s2 Submit Part B Find the direction of the acceleration of the particle. View Available Hint(s) The acceleration of the particle is toward the 8.75 kg mass. The acceleration of the particle is toward the 12.0 kg mass.

Answer 1

0 to F₂ mu 8.75Kg 18cm - 32cm 12 kg so cm Gravitation fouile on m due to 8.75kg mau FI= Gm x 8,75 270,061Gm (0.182 Similarly, F2= Gm x 12 117,187 Gm (0, 322 As Fix Fz iso the net force will be in the dirn of Fi. Net force Fhet Fir F2 = (270,061 - 117, 187) Gm acceleration : a = Fhet (270.061 – 11 71189) GyT ИЛ ty a = 152,874 x 6.67X10 = 102x10-8 m/s2 = As the met force acts along Fl lie tow ands 8,75 Kg mass. so the dirn of acceleration of the particle is also towards the 8,75 kg may.