Question # 1 Marks = 12

A worker pushes 25Kg crate a distance of 5m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic frication between the crate and the floor is 0.23.
(a) What magnitude of force must the worker apply?
(b) How much work is done on the crate by this force?
(c) How much work is done on the crate by frication?
(d) How much work is done on the crate by normal force?
(e) How much work is done on the crate by the gravity?
(f) What is the total work done on the crate?

Wc = mg = 25kg * 9.8N/kg = 245N. = Weight of crate.

Fc = 245N @ 0deg.
Fp = 245sin(0) = 0 = Force parallel to floor.
Fv = 245cos(0) = 245N. = Force perpendicular to floor = Normal.

Ff = u*Fv = 0.23 * 245 = 56.35N. = Force of friction.

a. Fn = Fap - Fp - Ff = ma = 0, a = 0.
Fap - 0 - 56.35 = 0,
Fap = 56.35N. = Force applied.

b. W = Fap*d = 56.35 * 5 = 281.8J.

c. W = Ff*d = 56.35 * 5 = 281.8J.

d. W = Fv*h = 245 * 0 = 0.

e. W = Fc*h = 250 * 0 = 0.

you're way smarter than I am.

To find the answers to the questions, we need to use the following formulas:

(a) The force required to push the crate can be calculated using the equation:
F = μ * N
where F is the force, μ is the coefficient of kinetic friction, and N is the normal force acting on the crate.

(b) The work done on the crate by the worker can be calculated using the equation:
W = F * d
where W is the work done, F is the magnitude of force, and d is the distance.

(c) The work done on the crate by friction can be calculated using the equation:
W_friction = μ * N * d
where W_friction is the work done by friction, μ is the coefficient of kinetic friction, N is the normal force acting on the crate, and d is the distance.

(d) The work done on the crate by the normal force is zero. This is because the normal force is perpendicular to the direction of displacement, so there is no work done.

(e) The work done on the crate by gravity can be calculated using the equation:
W_gravity = m * g * h
where W_gravity is the work done by gravity, m is the mass of the crate, g is the acceleration due to gravity, and h is the height.

(f) The total work done on the crate is the sum of the work done by the worker, friction, and gravity.

Now let's calculate each part:

(a) To find the magnitude of force, we need to calculate the normal force first. The normal force is equal to the weight of the crate, which can be calculated using the equation:
N = m * g
where N is the normal force, m is the mass of the crate, and g is the acceleration due to gravity.
Given, mass (m) = 25 kg and acceleration due to gravity (g) = 9.8 m/s^2.

N = 25 kg * 9.8 m/s^2 = 245 N
Now, substitute the value of the normal force into the equation for force:
F = μ * N = 0.23 * 245 N = 56.35 N

Therefore, the magnitude of force required to push the crate is 56.35 N.

(b) The work done on the crate by the worker can be calculated using the equation:
W = F * d
Given, distance (d) = 5 m and force (F) = 56.35 N.

W = 56.35 N * 5 m = 281.75 J

Therefore, the work done on the crate by the worker is 281.75 J.

(c) The work done on the crate by friction can be calculated using the equation:
W_friction = μ * N * d
Given, distance (d) = 5 m, coefficient of kinetic friction (μ) = 0.23, and normal force (N) = 245 N.

W_friction = 0.23 * 245 N * 5 m = 281.75 J

Therefore, the work done on the crate by friction is 281.75 J.

(d) The work done on the crate by the normal force is zero.

(e) The work done on the crate by gravity can be calculated using the equation:
W_gravity = m * g * h
Since the crate is moved along a level floor, the height (h) is zero. Therefore, the work done by gravity is zero.

(f) The total work done on the crate is the sum of the work done by the worker, friction, and gravity:
Total work = Work by the worker + Work by friction + Work by gravity
= 281.75 J + 281.75 J + 0 J
= 563.5 J

Therefore, the total work done on the crate is 563.5 J.