Answer:
the work done by the worker in pulling the crate is 500 J, and the kinetic energy of the crate at the end of the 10 m distance is 292.13 J.
Step-by-step explanation:
To solve the problem, we need to first resolve the applied force and the frictional force into their horizontal and vertical components. The horizontal component of the applied force is:
F_h = Fcos(30°) = 50cos(30°) = 43.3 N
The vertical component of the applied force is:
F_v = Fsin(30°) = 50sin(30°) = 25 N
The frictional force is acting in the opposite direction to the applied force, so its horizontal component is:
f_h = -F_f = -20 N
Since the crate is initially at rest, the net force on the crate is equal to the applied force minus the frictional force:
F_net = F_h + f_h = 43.3 - 20 = 23.3 N
The acceleration of the crate is given by Newton's second law:
F_net = ma
where a is the acceleration of the crate. Rearranging this equation, we get:
a = F_net/m = 23.3/30 = 0.78 m/s²
The work done by the worker in pulling the crate a distance of 10 m is given by:
W = Fdcos(θ)
where d is the distance pulled, and θ is the angle between the applied force and the displacement. In this case, θ = 0° since the force is applied horizontally. Therefore, the work done is:
W = Fd = 5010 = 500 J
The kinetic energy of the crate at the end of the 10 m distance is:
K = (1/2)mv²
where v is the final velocity of the crate. We can find v using the equation of motion:
v² = u² + 2as
where u is the initial velocity (zero), s is the displacement (10 m), and a is the acceleration (0.78 m/s²). Therefore:
v² = 0 + 20.7810 = 15.6
v = sqrt(15.6) = 3.95 m/s
Substituting this value of v into the equation for kinetic energy, we get:
K = (1/2)mv² = (1/2)30(3.95)² = 292.13 J
Therefore, the work done by the worker in pulling the crate is 500 J, and the kinetic energy of the crate at the end of the 10 m distance is 292.13 J.