Consider two masses m1 and m2 with m2=2m1, both masses are pushed a distance d b
ID: 2142628 • Letter: C
Question
Consider two masses m1 and m2 with m2=2m1, both masses are pushed a distance d by the same constant force F,
1) is the work done by the force in case 1 greater than, less than or equal to the work done by the force in 2?
2) is the change in kinetic energy of m2 in case 1 greater than, less then or equal to the change in kinetic energy of m2?
3)is the final velocity of m1 in case 1 greater than, less than or equal to the final velocity of m2 in case 2?
Explanation / Answer
1). work done = force*displacement work done by the force in case 1 is equal to the work done by the force in 2. 2). work done = change in kinetic energy change in kinetic energy of m2 in case 1 is equal to the change in kinetic energy of m2. 3). vi1 = vi2 = 0 (since the masses are at rest initially) work done in case 1 = 0.5*m1*[vf1^2 - vi1^2] = 0.5*m1*vf1^2 work done in case 2 = 0.5*m2*[vf2^2 - vi2^2] = 1*m1*[vf2^2 - vi2^2] = m1*vf2^2 work done in case 1 = work done in case 2 => 0.5*m1*vf1^2 = m1*vf2^2 => (vf1/vf2)^2 = 2 vf1 is greater than vf2. final velocity of m1 is greater than final velocity of m2.