Force of Friction: 0.45N, Coefficient: 0.0382

[italystaly]

A force of 1.2N is applied to an object of mass 1.5Kg. It accelerates at 0.50 m/s^2. Determine the Force of friction and the Coefficient of Friction

This is what I did. However, it seems that the Force of Friction is really low so I'm not sure if I did this question right.
F=M * A ..which is can also be used as Fnet
.:. Fnet= 1.5kg(0.50 m/s^2)
= 0.75 N

Fapp= 1.2 N

Fnet=Fapplied + Ffriction
Ffriction = Fnet-Fapplied
= 0.75n -1.2 n
= - 0.45 N


If I continue to Figure out the coefficient of Friction it seems very unreasonable.

Ffriction/ Fnormal = kinetic Coefficient

-.45n/ -9.8m/s^2 * 1.5kg
= .0382



This may be right but I'm not quite sure. Any input would be great.

 

[kuruman]

Draw a free body diagram. Say the applied force is to the right. Then the force of friction must be to the left. So from the FBD you get

Fnet=Fapplied - Ffriction, not what you have.

 

[kerimek]

I would like to clarify a few things about the given information. The force of friction is a resistive force that opposes the motion of an object. In this case, the given value of 0.45N may seem low, but it is important to note that this force is dependent on the coefficient of friction and the normal force, not just the mass of the object. The coefficient of friction, represented by the Greek letter mu (μ), is a dimensionless quantity that describes the roughness of the surfaces in contact and their tendency to stick or slide against each other.

To determine the force of friction, we need to first find the normal force acting on the object. This is equal to the weight of the object, which can be calculated as mass x acceleration due to gravity (9.8 m/s^2). Therefore, the normal force in this case would be 1.5kg x 9.8 m/s^2 = 14.7N.

Now, using the formula Ffriction = μ x Fnormal, we can calculate the force of friction as 0.0382 x 14.7N = 0.561N. This value is higher than the given force of 0.45N, which may indicate that the coefficient of friction is not accurate or that there are other forces at play.

Furthermore, the given information does not specify whether the applied force of 1.2N is the net force or the total force. If it is the net force, then the acceleration of 0.50 m/s^2 would be incorrect as it should be calculated using the total force. Without this clarification, it is not possible to accurately determine the force of friction and the coefficient of friction.

In conclusion, as a scientist, I would advise gathering more information and clarifying the given values before attempting to solve this problem. It is important to always consider all the relevant factors and use accurate and precise values in scientific calculations.