Answer:
See answer in the detailed solution
Explanation:
Exact statement of question is :
A boy drags a wooden crate with a mass of 20.0 kg, a distance of 12.0 m, across a rough level floor
at a constant speed of 1.50 m/s by pulling on the rope tied to the crate with a
force of 50.0 N. The
rope makes an angle of 25.0 ° with the horizontal.
a. What are the horizontal and vertical components of the applied force?
b. What is the magnitude of each of the forces?
Applied=
Weight =
Normal =
Frictional=
c) How much work is done by each of the forces?
d. What is the total amount of work done on the crate?
e. What is the coefficient of friction of the crate on the floor?
Solution:
Fx= F cos 25
= 50 × 0.91 = 45.5 N
and
Fy = F sin 25 = 50 × 0042261 = 21 N
So we can say;
1- applied force = 50 N
2- Weight = mg = 20×10= 200 N
3- Normal force = weight - normal component of applied force = 200-21=179 N
4- Force of friction is equal to horizontal component of applied force = 45.5 N
5- Work done by weight force and normal forces is zero - because forces are making 90 degree with direction of movement.
6- Work done by friction force = distance × force = 12 × 45.5 = 546 J
7- Total work done on box is zero because it moves with constant speed.
8- C0-efficient of friction μ = [tex]\frac{Fx}{W-Fy}[/tex]
[tex]=\frac{45.5}{200-21}[/tex]
==> μ = 0.25
The magnitude of tension force acting on the crate is 50cos(25) = 45.31 N.
And the magnitude of frictional force acting is 45.31 N.
The rope is inclined at 25° from the horizontal. Therefore the magnitude of the force acting on the crate will be the cosine of actual tension in the rope, which will be cos25° times 50N (given).
F(t) = 50cos25° = 45.31 N
Since the crate is moving with a constant velocity, there is no acceleration, which means there is no net force acting on the crate.
We can conclude that the magnitude of frictional force f is equal to F(t).
f = F(t) = 45.31 N
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