A 20 kg wooden crate is dragged 12 m across a rough level floor by a rope. The force of tension in the rope is 50 N at an angle of 25° above the horizontal and the crate is moving at a constant speed of 1.50 m/s. Find the magnitude of tension force acting on the crate in [N].

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 =

==> μ = 0.25

yeet youwelcome broo

0.11 m/s and 0.28 m/s should be correct