Question

A 20 g fridge magnet is being held onto a fridge by a 0.9 N force. Assuming μs = 0.30, calculate: A. Force of friction acting on the magnet B. The weight of the magnet C. The acceleration of the magnet

Answer 1

The force of friction acting on the magnet is 0.9 N. The weight of the 20 g magnet is 0.196 N. Since the magnet is static, the acceleration is 0 m/s².

Step-by-step explanation:

The question involves calculating the force of friction, the weight of the magnet, and the acceleration of the magnet using provided values and the concept of static friction. To calculate these, use the following formulas:

• Force of friction (f) is given by f = μs × N, where μs is the coefficient of static friction and N is the normal force.
• The weight (W) of an object is calculated as W = m × g, where m is the mass of the object and g is the acceleration due to gravity.
• The acceleration (a) can be found using Newton's second law if the net force and mass are known: a = F/m, where F is the net force acting on the object.

A. The force of friction acting on the magnet can be found using the coefficient of static friction (μs) and the normal force, which is equal to the weight of the magnet. Since we're dealing with a static situation and the magnet is not moving, the force of friction is equal to the force holding the magnet against the fridge, thus f = 0.9 N.

B. The weight of the magnet (W) can be calculated using W = m × g. Given the mass (m) is 20 g, which needs to be converted to kilograms (0.020 kg), and the acceleration due to gravity (g) is approximately 9.8 m/s², the weight W = 0.020 kg × 9.8 m/s² = 0.196 N.

C. As the magnet is not moving (it is static), the acceleration of the magnet is 0 m/s² because there is no net force causing a change in velocity.

Answer 2

Given,

The mass of the fridge magnet, m=20 g

The force that is holding the fridge magnet, F=0.9 N

The coefficient of friction, μ_s=0.30

Where N is the normal force acting on the fridge magnet and f is the frictional force.

A.

The normal force acting on the fridge magnet is equal to the force that holds the fridge magnet to the surface of the fridge.

Thus the normal acting on fridge magnet is,

`N=F=0.9\text{ N}`

Thus the frictional force acting on the magnet is given by,

`\begin{gathered} f=N\mu \\ \end{gathered}`

On substituting the known values,

`\begin{gathered} f=0.9*0.3 \\ =0.27\text{ N} \end{gathered}`

Thus the frictional force acting on the magnet is 0.27 N

B.

The weight of the magnet is given by,

`W=mg`

Where g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} W=20*9.8 \\ =196\text{ N} \end{gathered}`

Thus the weight of the magnet is 196 N

C.

As the friction acting on the magnet is less than the weight of the magnet, the magnet will fall down.

The net vertical force acting on the magnet is given by,

`\begin{gathered} F_n=ma \\ =W-f \end{gathered}`

Where a is the acceleration of the magnet.

On substituting the known values,

`\begin{gathered} 20* a=196-0.27_{} \\ a=(195.73)/(20) \\ =9.77\text{ m/s}^2 \end{gathered}`

Thus the acceleration of the magnet is 9.77 m/s²