Another Question: Find the work done by the kinetic frictional force

AI Thread Summary
The discussion focuses on calculating the work done by the kinetic frictional force on a refrigerator being pulled across a surface. The correct formula for work is W = Fd cosθ, where F is the frictional force, d is the distance, and θ is the angle between the force and displacement. The kinetic frictional force is determined using Ff = μN, with N being the normal force equal to the weight of the refrigerator. After calculations, the work done by the kinetic frictional force is found to be -1058.4 J, indicating it acts in the opposite direction of movement. The total work done on the refrigerator, considering both the pulling force and friction, is 611.6 J.
zizikaboo
Messages
3
Reaction score
0
A 300N force is pulling an 90.0 kg refrigerator across a horizontal surface. The force acts at an angle of 22.0° above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 6.00 m.

i found out the work done by pulling force is 1669 J and its asking for the work done by the kinetic frictional force!
here's what i did but the answer is wrong though...i tried a million times and it's still wrong! i know something's wrong here and i know it's not the acceleration of gravity or m...hmm

i did
mg times mu=9.8*90*.2
then i times it by the distance to get work. 176.4N*6m=1058.4 J
which is WRONG! arghh...am i missing something here? :smile:
 
Physics news on Phys.org
zizikaboo said:
A 300N force is pulling an 90.0 kg refrigerator across a horizontal surface. The force acts at an angle of 22.0° above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 6.00 m.


i did
mg times mu=9.8*90*.2
then i times it by the distance to get work. 176.4N*6m=1058.4 J
which is WRONG!

The friction is equal the normal force multiplied by the coefficient of friction.

F_{fr} = \mu F_N

As the refrigerator slides on a horizontal plane the vertical component of the resultant force is zero. This vertical component is the normal force + vertical component of the pulling force - mg.

F_N+ F\sin (\alpha) - mg = 0.

ehild
 


The correct way to calculate the work done by the kinetic frictional force is to use the formula W = Fd cosθ, where W is the work done, F is the force, d is the distance, and θ is the angle between the force and the displacement.

In this case, the force acting against the movement of the refrigerator is the kinetic frictional force, which is given by Ff = μN, where μ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the refrigerator, so N = mg.

Plugging in the given values, we get Ff = (0.200)(90 kg)(9.8 m/s^2) = 176.4 N. Now, we can calculate the work done by the kinetic frictional force using the formula: W = Ff d cosθ = (176.4 N)(6.00 m)cos(180°) = -1058.4 J.

Note that the work done by the kinetic frictional force is negative because it is acting in the opposite direction of the displacement. This means that the kinetic frictional force is doing negative work, which is taking away energy from the system. Therefore, the total work done by all forces on the refrigerator is the sum of the work done by the pulling force (1669 J) and the work done by the kinetic frictional force (-1058.4 J), which is equal to 611.6 J.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top