1. A boy drags a wooden crate with a mass of 20 kg, a distance of 12 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 N. The rope makes an angle of 25¡ã with the horizontal.
a.What is the magnitude of each of the forces?
b.How much work is done by each of the forces?
c.What is the total amount of work done on the object?
d.What is the coefficient of friction of the crate on the floor? (Note, we usually write coefficients of friction to only 2 digits of accuracy.)
Wc = mg = 20kg * 9.8N/kg=196N.= Weight of crate.
a. Fc = [196N,0deg).
Fp = 196sin(0) = 0 = Force parallel to
floor.
Fv = 196cos(0) = 196N. = Force perpendicular to floor.
Fn = Fap*cosA - Fp - Ff = 0,
Fn = 50cos25 - 0 - Ff = 0,
50cos25 - Ff = 0,
Ff = 50cos25 = 45.32N. = Force of friction.
b. W = Fd.
W = 50*cosw25 * 12 = 544N.
W = Fp*d = 0 * 12 = 0.
W = Ff*d = 45.32 * 12 = 544N.
c. Wt = 544 + 544 = 1088N.
d. u = Ff / Fv = 45.32 / 196 = 0.23.
Correction: All work is in Joules and NOT Newtons.
Step 1: Find the horizontal and vertical components of the force applied to the crate.
The horizontal component of the force:
F_horizontal = F * cos(angle)
F_horizontal = 50 N * cos(25°)
F_horizontal = 50 N * 0.9063
F_horizontal ≈ 45.32 N
The vertical component of the force:
F_vertical = F * sin(angle)
F_vertical = 50 N * sin(25°)
F_vertical = 50 N * 0.4236
F_vertical ≈ 21.18 N
a. The magnitude of each force:
The magnitude of the force applied to the crate is 50 N, and the magnitude of the force opposing the motion (friction) is equal to the horizontal component of the applied force, which is approximately 45.32 N.
b. The work done by each force:
The work done by the applied force is given by the equation:
Work = force * distance * cos(angle)
Work_applied = F * d * cos(angle)
Work_applied = 50 N * 12 m * cos(25°)
Work_applied ≈ 611.89 J
The work done by the friction force is given by the equation:
Work = force * distance * cos(180°) = -force * distance
Work_friction = -F_horizontal * d
Work_friction = -45.32 N * 12 m
Work_friction ≈ -543.84 J (the negative sign indicates that it is work done against the direction of motion)
c. The total amount of work done on the object:
The total work done on the object is the sum of the work done by the applied force and the work done by the friction force:
Total work done = Work_applied + Work_friction
Total work done = 611.89 J + (-543.84 J)
Total work done ≈ 68.05 J
So, the total amount of work done on the object is approximately 68.05 J.
d. The coefficient of friction:
The coefficient of friction can be found by dividing the frictional force by the normal force.
Frictional force = F_horizontal = 45.32 N
The normal force is equal to the weight of the crate:
Normal force = mass * gravitational acceleration
Normal force = 20 kg * 9.8 m/s²
Normal force = 196 N
Coefficient of friction = Frictional force / Normal force
Coefficient of friction = 45.32 N / 196 N
Coefficient of friction ≈ 0.231 (rounded to 2 decimal places)
Therefore, the coefficient of friction of the crate on the floor is approximately 0.23.
a. To find the magnitude of each force, we can break down the forces into their horizontal and vertical components. The horizontal component of the force exerted by the boy can be found using the equation Fx = F * cos(theta), where F is the magnitude of the force (50 N) and theta is the angle (25 degrees).
Fx = 50 N * cos(25 degrees)
≈ 45.98 N
The vertical component of the force exerted by the boy can be found using the equation Fy = F * sin(theta), where F and theta are the same as above.
Fy = 50 N * sin(25 degrees)
≈ 21.29 N
The vertical component of the force is balanced by the normal force acting on the crate, so the magnitude of the normal force is also equal to 21.29 N.
b. The work done by each force can be calculated using the formula W = F * d * cos(theta), where W is the work done, F is the magnitude of the force, d is the distance, and theta is the angle between the force and the displacement.
For the force exerted by the boy:
W = 50 N * 12 m * cos(25 degrees)
≈ 555.2 J
For the normal force:
W = 21.29 N * 12 m * cos(0 degrees)
= 0 J (since the normal force is perpendicular to the displacement)
c. The total amount of work done on the object is the sum of the work done by each force.
Total work = Work by the boy's force + Work by the normal force
≈ 555.2 J + 0 J
≈ 555.2 J
d. To find the coefficient of friction, we can use the formula f = μN, where f is the magnitude of the frictional force, μ is the coefficient of friction, and N is the magnitude of the normal force.
Since the crate is moving with a constant speed, the frictional force must be equal in magnitude and opposite in direction to the force exerted by the boy.
Frictional force = Force exerted by the boy
= 45.98 N
Using the equation above, and substituting the magnitude of the normal force from part a, we get:
45.98 N = μ * 21.29 N
Simplifying, we find:
μ ≈ 2.16
Therefore, the coefficient of friction of the crate on the floor is approximately 2.16 (to 2 decimal places).