Respuesta :
Answer:
a. 900 J
b. 0.383
Explanation:
According to the question, the given data is as follows
Horizontal force = 150 N
Packing crate = 40.0 kg
Distance = 6.00 m
Based on the above information
a. The work done by the 150-N force is
[tex]W = F x = \mu N x = \mu\ m\ g\ x[/tex]
[tex]W = 150 \times 6[/tex]
= 900 J
b. Now the coefficient of kinetic friction between the crate and surface is
[tex]\mu = \frac {F}{m\timesg}[/tex]
[tex]= \frac{150}{40\times 9.8}[/tex]
= .383
We simply applied the above formulas so that each one part could calculate
We want to find the work and kinetic friction for the given situation. The solutions are:
- a) W = 900 N*m
- b) μ = 0.38
Here we have a horizontal force of 150N pushing a 40.0 kg packing crate a distance of 6.00m at a constant speed.
a) First we want to find the work, it is given by the force applied times the distance moved, so the work is just:
W = 150N*6.00m = 900 N*m
b) Now we want to find the coefficient of kinetic friction, it must be such that the kinetic friction force is equal to the pushing force, in this way there is no net force, and then there is no acceleration.
Remember that the friction force is:
F = m*g*μ
Where:
- m = mass of the box = 40 kg
- g = gravitational acceleration = 9.8m/s^2
- μ = coefficient of kinetic friction.
Then we must solve:
150N = 40kg*(9.8 m/s^2)*μ = 392N*μ
150N/392N = 0.38
So the coefficient of kinetic friction between the crate and the surface is 0.38
If you want to learn more about forces, you can read:
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