Sally applies a total force of 71 N with a rope
to drag a wooden crate of mass 100 kg across
a floor with a constant acceleration in the x-
direction of a = 0.1 m/s2. The rope tied to
the crate is pulled at an angle of � = 50 �
relative to the floor.
Calculate the total work done by friction as
the block moves a distance 73.7 m over the
horizontal surface. The acceleration due to
gravity is 9.8 m/s2 .
Answer in units of J
Well, Sally seems to have quite the pulling power! Now, let's get down to business and calculate the total work done by friction.
The force Sally applies can be divided into two components: one in the horizontal direction and one in the vertical direction.
The horizontal component of the force is given by:
F_horizontal = F * cos(theta)
= 71 N * cos(50 degrees)
F_horizontal ≈ 71 N * 0.6428 ≈ 45.57 N
The frictional force, which opposes the motion, will be equal in magnitude to the horizontal component of the applied force (since there is no vertical motion), but in the opposite direction:
Frictional force = - 45.57 N
Now, we can calculate the work done by friction using the formula:
Work = force * distance * cos(theta)
Since the force and the distance are in the same direction (as the block is moving horizontally), the angle between them is 0 degrees. Therefore, cos(0) = 1.
Work = frictional force * distance * cos(0)
= - 45.57 N * 73.7 m * cos(0)
Given that cos(0) = 1, we can simplify this to:
Work = - 45.57 N * 73.7 m
Now, let's calculate the final answer:
Work ≈ - 3358.749 J
Since work is a scalar quantity, the negative sign just represents the direction of the force opposing motion. Therefore, the total work done by friction is approximately 3358.749 J.
To find the total work done by friction, we first need to find the force of friction acting on the crate. The force of friction can be calculated using the equation:
Friction = μ * Normal Force
Here, μ is the coefficient of friction and the normal force is the force exerted by the floor on the crate, which is equal to the weight of the crate.
The weight of the crate can be calculated using the equation:
Weight = mass * acceleration due to gravity
Let's calculate the weight first:
Weight = 100 kg * 9.8 m/s^2
Weight = 980 N
Now, we need to find the coefficient of friction. The force of friction can be determined using the equation:
Friction = force applied * sin(θ) - force applied * cos(θ) * μ
Here, θ is the angle at which the force is applied (50 degrees), and force applied is the total force applied by Sally through the rope (71 N).
Friction = 71 N * sin(50 degrees) - 71 N * cos(50 degrees) * μ
To solve for μ, we need to rearrange the equation:
μ = (71 N * sin(50 degrees) - Friction) / (71 N * cos(50 degrees))
Now, let's calculate the coefficient of friction using the given values:
μ = (71 N * sin(50 degrees) - Friction) / (71 N * cos(50 degrees))
Next, let's find the frictional force using the equation:
Friction = μ * Normal Force
Friction = μ * Weight
Finally, we can calculate the work done by friction using the equation:
Work = Friction * distance
Given that the distance is 73.7 m, we can now calculate the work done by friction.
To calculate the total work done by friction on the wooden crate, we need to first determine the net force acting on the crate.
The net force is given by the equation:
Net Force = mass x acceleration
In this case, the mass of the crate is 100 kg and the acceleration is 0.1 m/s^2.
Net Force = 100 kg x 0.1 m/s^2 = 10 N
The net force is in the opposite direction to the force applied by Sally with the rope. So, the force of friction can be calculated as:
Force of Friction = Force Applied - Net Force
Force Applied by Sally = 71 N (given in the question)
Force of Friction = 71 N - 10 N = 61 N
The work done by friction is given by the equation:
Work = Force x Distance x cos(theta)
In this case, the distance is given as 73.7 m and the angle theta is given as 50 degrees. However, the angle theta should be converted to radians before using it in the equation.
Radians = Degrees x (π/180)
theta in radians = 50 degrees x (π/180) ≈ 0.8727 radians
Work = 61 N x 73.7 m x cos(0.8727)
So, the total work done by friction as the block moves a distance of 73.7 m is:
Work = 61 N x 73.7 m x cos(0.8727) ≈ 4114 Joules (J)
Fap = 71N. @ 50o.
Wc = m*g = 100kg * 9.8N/kg = 980 N. = Wt. of crate.
Fc = 980N. @ 0o. = Force of crate.
Fp = 980*sin(0) = 0 = Force parallel to
floor.
Fv = 980*cos(0) = 980 N. = Force perpendicular to floor.
Fn = Fap*cosA-Fp-Fk = ma.
71*cos50-0-Fk = 100*0.1 = 10
-Fk = 10 - 71*cos50 = -35.63
Fk = 35.63 N.=Force of kinetic friction.
Work = Fk*d = 35.63 * 73.7 = 2700 Joules