Question

A very large magnet applies a repulsive force to a smaller (0.05 Kg) magnet. If the smaller magnet accelerates across a frictionless table at 2m/s2, then what is the magnitude of the repulsive force?​

Answer 1

Repulsive force = 0.1 N

Step-by-step explanation:

Given:

Mass of the smaller magnet is,
`m=0.05\ kg`

Acceleration of the smaller magnet is,
`a=2\ m/s^2`

Frictional force is zero as the table is frictionless.

From Newton's second law, we know that the net force acting on an object is equal to the product of its mass and the net acceleration. The acceleration of the object acts in the direction of the net force.

Here, the mass and net acceleration acting on the magnet is given. So, we use Newton's second law to find repulsive force acting on it.

Therefore, from Newton's second law, the net force acting on the magnet is only the repulsive force and is given as:

`F_(rep)=ma`

Plug in 0.05 kg for 'm', 2 m/s² for 'a' and solve for
`F_(rep)`. This gives,

`F_(rep)=(0.05\ kg)(2\ m/s^2)\\F_(rep)=0.1\ kgms^(-2)\\F_(rep)=0.1\ N`

Therefore, the repulsive force acting on the smaller magnet by the larger magnet is 0.1 N.