A black aluminumglider floats on a film of air above a level aluminum air track.

Question

A black aluminumglider floats on a film of air above a level aluminum air track.Aluminum feels essentially no force in a magnetic field, and airresistance is negligible. A strong magnet is attached to the top ofthe glider, forming a total mass of 240 g. A piece of scrapiron attached to one end stop on the track attracts the magnet witha force of 0.823 N when the iron and the magnet are separated by2.50cm. (a) Find the acceleration of the glider at this instant. (b)The scrap iron is now attached to another green glider,forming a total mass of 120 g. Find the acceleration of each gliderwhen they are simultaneously released at 2.50 cmseparation.
A black aluminumglider floats on a film of air above a level aluminum air track.Aluminum feels essentially no force in a magnetic field, and airresistance is negligible. A strong magnet is attached to the top ofthe glider, forming a total mass of 240 g. A piece of scrapiron attached to one end stop on the track attracts the magnet witha force of 0.823 N when the iron and the magnet are separated by2.50cm. (a) Find the acceleration of the glider at this instant. (b)The scrap iron is now attached to another green glider,forming a total mass of 120 g. Find the acceleration of each gliderwhen they are simultaneously released at 2.50 cmseparation.
(a) Find the acceleration of the glider at this instant. (b)The scrap iron is now attached to another green glider,forming a total mass of 120 g. Find the acceleration of each gliderwhen they are simultaneously released at 2.50 cmseparation.

Explanation / Answer

The total mass of the strong magnet attached to the top of theglider is m = 240 g = 240 * 10-3 kg The piece of scrap iron attracts the magnet with a force of F= 0.823 N (a)Let the acceleration of the glider at this end bea.Therefore,we get F = m * a or 0.823 = 240 * 10-3 * a or a = (0.823/240 * 10-3) or a = 3.43 m/s2 (b)The total mass of the green glider is m1= 120 g= 120 * 10-3 kg Let the acceleration of each glider when they aresimultaneously released at 2.50 cm separation be a1 anda2. When the total mass of the glider attached to the strongmagnet is m = 240 g = 240 * 10-3 kg,then the forcebetween the scrap iron and the magnet is F = (GmM/r2) Here,G is the universal gravitational constant,m is the massof the scrap iron and M is the mass of the magnet.The value of G is6.67 * 10-11 Nm2/kg2 and r = 2.50cm = 2.50 * 10-2 m. or M * a1= (GmM/r2) or a1= (Gm/r2) Substituting the values in the above equation,we get a1= (6.67 * 10-11 * 240 *10-3/(2.50 * 10-2)2) or a1= 2.56 * 10-8 m/s2 When the scrap iron is attached to another greenglider,forming a total mass of m1= 120 g = 120* 10-3 kg,then the force between the scrap iron and themagnet is F = (Gm1M/r2) or M * a2= (Gm1M/r2) or a2= (Gm1/r2) Substituting the values in the above equation,we get a2= (6.67 * 10-11 * 120 *10-3/(2.50 * 10-2)2) or a2= 1.28 * 10-8 m/s2 Therefore,the acceleration of each glider when they aresimultaneously released at 2.50 cm separation is a1= 2.56 * 10-8 m/s2 anda2= 1.28 * 10-8 m/s2. (b)The total mass of the green glider is m1= 120 g= 120 * 10-3 kg Let the acceleration of each glider when they aresimultaneously released at 2.50 cm separation be a1 anda2. When the total mass of the glider attached to the strongmagnet is m = 240 g = 240 * 10-3 kg,then the forcebetween the scrap iron and the magnet is F = (GmM/r2) Here,G is the universal gravitational constant,m is the massof the scrap iron and M is the mass of the magnet.The value of G is6.67 * 10-11 Nm2/kg2 and r = 2.50cm = 2.50 * 10-2 m. or M * a1= (GmM/r2) or a1= (Gm/r2) Substituting the values in the above equation,we get a1= (6.67 * 10-11 * 240 *10-3/(2.50 * 10-2)2) or a1= 2.56 * 10-8 m/s2 When the scrap iron is attached to another greenglider,forming a total mass of m1= 120 g = 120* 10-3 kg,then the force between the scrap iron and themagnet is F = (Gm1M/r2) or M * a2= (Gm1M/r2) or a2= (Gm1/r2) Substituting the values in the above equation,we get a2= (6.67 * 10-11 * 120 *10-3/(2.50 * 10-2)2) or a2= 1.28 * 10-8 m/s2 Therefore,the acceleration of each glider when they aresimultaneously released at 2.50 cm separation is a1= 2.56 * 10-8 m/s2 anda2= 1.28 * 10-8 m/s2. Therefore,the acceleration of each glider when they aresimultaneously released at 2.50 cm separation is a1= 2.56 * 10-8 m/s2 anda2= 1.28 * 10-8 m/s2.

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