Two objects of different mass start from rest, are pulled by the same magnitude net force, and are moved through the same distance. The work done on object A is 500J. After the force pulled each object, object A moves twice as fast as object B. Answer the following:

a.) How much work is done on object B

b.) What is the kinetic energy of object A after being pulled?

c.) What is the kinetic energy of object B after being pulled?

d.) What is the ratio of the mass of object A to the mass of object B?

250

a) 500J

b) 500J
c) 500J
d) 1:2

To solve this problem, we can use the work-energy principle and the equation for kinetic energy.

a.) To find the work done on object B, we know that work is equal to force times distance. Since both objects were pulled by the same magnitude net force and moved through the same distance, the work done on object B will also be 500J.

b.) The kinetic energy of an object can be calculated using the formula: KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass, and v is the velocity. We are given that object A moves twice as fast as object B.

Let's assume the mass of object A is mA and the velocity of object B is vB. Therefore, the velocity of object A will be 2vB. We are also given that the work done on object A is 500J.

Since work is equal to the change in kinetic energy, we can write:

500J = 1/2 * mA * (2vB)^2

Simplifying the equation:

500J = 1/2 * mA * 4vB^2

500J = 2 * 1/2 * mA * vB^2

500J = 1 * mA * vB^2

500J = mA * vB^2

Divide both sides by vB^2:

500J / vB^2 = mA

c.) The kinetic energy of object A after being pulled can be found using the equation mentioned earlier. We can now substitute the values we have:

KE of A = 1/2 * mA * (2vB)^2

Simplifying:

KE of A = 1/2 * mA * 4vB^2

KE of A = 2 * 1/2 * mA * vB^2

KE of A = 1 * mA * vB^2

Since we know from part b that 500J = mA * vB^2, we can substitute this value:

KE of A = 500J

So, the kinetic energy of object A after being pulled is 500J.

d.) To find the ratio of the mass of object A to the mass of object B, we can divide the equation we found in part b by the equation we found in part c:

500J / 500J = mA / mB

Therefore, the ratio of the mass of object A to the mass of object B is 1:1 or mA = mB.

a.) Well, since object A moves twice as fast as object B, it means that object A is putting on a real show! You could say object A is the Usain Bolt of objects! And since both objects are pulled by the same magnitude net force and move through the same distance, that means the work done on object B should be the same as object A. So, the work done on object B is also 500J.

b.) After being pulled, object A must have some serious energy! It's like a cheetah on steroids! The kinetic energy of object A can be calculated using the formula KE = 0.5 * mass * velocity^2. But since we know that object A is moving twice as fast as object B, we can say that its velocity is twice that of object B. So, the kinetic energy of object A is 4 times the kinetic energy of object B.

c.) Now, for object B, its kinetic energy might not be as impressive as A's, but it still has some moves! Using the same formula, we can calculate the kinetic energy of object B. Since we know the kinetic energy of object A is 4 times that of object B, we can say that the kinetic energy of object B is 1/4th the kinetic energy of object A.

d.) Ah, the ratio of the masses! That's a piece of cake! Since we know the work done on object A and B is the same, we can use the formula for work done, which is W = force * distance. The distance and force are the same for both objects, so that means the ratio of the mass of object A to the mass of object B is also 1:1. It's a balanced act, just like a seesaw with equal weights on each side!