Can anyone explain this to me, it was on my exam and i don't get it ;
A 14.0 kg crate is pulled up a rough incline with an initial speed of 1.6 m/s. The pulling force is 131.0 N parallel to the incline, which
makes an angle of 12.3 ◦ with the horizontal. The coefficient of kinetic friction is 0.26 and the crate is pulled a distance of 8.4 m. The acceleration of gravity is 9.81 m/s
a) Find the work done by Earth’s gravity
on the crate.
Answer in units of J.
b) Find the work done by the force of friction
on the crate.
Answer in units of J
c) Find the work done by the puller on the
crate.
Answer in units of J.
d) Find the change in kinetic energy of the
crate.
Answer in units of J
e) Find the speed of the crate after it is pulled 8.4 m.
Answer in units of m/s.
Sure! Let's break down the problem and solve it step-by-step.
Given:
Mass of crate (m) = 14.0 kg
Initial speed (v₀) = 1.6 m/s
Pulling force (F) = 131.0 N
Angle of incline (θ) = 12.3°
Coefficient of kinetic friction (μk) = 0.26
Distance (d) = 8.4 m
Acceleration due to gravity (g) = 9.81 m/s²
a) Work done by Earth's gravity on the crate:
The work done by gravity is calculated by the formula W = m * g * h, where h is the vertical distance traveled. In this case, the crate is pulled along the incline, so we need to find the vertical distance from the initial position to the final position.
Using trigonometry, the vertical height (h) can be calculated as h = d * sin(θ). Substituting the values:
h = 8.4 m * sin(12.3°) = 1.906 m
Now, we can calculate the work done by gravity:
W = m * g * h
W = 14.0 kg * 9.81 m/s² * 1.906 m
W ≈ 261.16 J
Therefore, the work done by Earth's gravity is approximately 261.16 J.
b) Work done by the force of friction on the crate:
The work done by the force of friction can be calculated using the formula W = μk * m * g * d, where d is the horizontal distance.
Substituting the values:
W = 0.26 * 14.0 kg * 9.81 m/s² * 8.4 m
W ≈ 288.2688 J
Therefore, the work done by the force of friction is approximately 288.2688 J.
c) Work done by the puller on the crate:
The work done by the puller is given by the product of the applied force and the distance moved in the direction of the force.
W = F * d * cos(θ)
W = 131.0 N * 8.4 m * cos(12.3°)
W ≈ 1064.65 J
Therefore, the work done by the puller on the crate is approximately 1064.65 J.
d) Change in kinetic energy of the crate:
The change in kinetic energy can be found using the work-energy theorem. It states that the work done on an object is equal to the change in its kinetic energy.
W = ΔKE
ΔKE = W
From part (c), we found that the work done by the puller is 1064.65 J.
Therefore, the change in kinetic energy of the crate is approximately 1064.65 J.
e) Speed of the crate after it is pulled 8.4 m:
To find the final speed, we can use the work-energy theorem again.
ΔKE = KEf - KEi
KEf = ΔKE + KEi
From part (d), we found that the change in kinetic energy is 1064.65 J. The initial kinetic energy (KEi) is given by (1/2) * m * v₀².
KEi = (1/2) * 14.0 kg * (1.6 m/s)² = 17.92 J
Now, calculating the final kinetic energy (KEf):
KEf = 1064.65 J + 17.92 J
KEf = 1082.57 J
To find the final speed (vf), we can use the equation KEf = (1/2) * m * vf² and solve for vf:
vf² = (2 * KEf) / m
vf² = (2 * 1082.57 J) / 14.0 kg
vf ≈ 10.04 m/s
Therefore, the speed of the crate after it is pulled 8.4 m is approximately 10.04 m/s.
Sure! Let's break down each part of the question and explain how to solve it.
a) The work done by Earth's gravity on the crate can be determined using the formula: Work = Force * Distance * cos(theta), where theta is the angle between the force and the direction of displacement. In this case, the force of gravity is the weight of the crate, which is given by the mass (14.0 kg) multiplied by the acceleration due to gravity (9.81 m/s^2). The distance is given as 8.4 m. Therefore, the work done by gravity can be calculated as:
Work_gravity = (14.0 kg * 9.81 m/s^2) * 8.4 m * cos(0°)
Work_gravity = 1152.10 J
So, the work done by Earth's gravity on the crate is 1152.10 J.
b) The work done by the force of friction can be calculated using the formula: Work = Force * Distance * cos(theta), where theta is the angle between the force and the direction of displacement. The force of friction can be determined by multiplying the coefficient of kinetic friction (0.26) by the normal force, which is the weight of the crate (mass * acceleration due to gravity). The distance is again given as 8.4 m. Therefore, the work done by friction can be calculated as:
Work_friction = (0.26 * (14.0 kg * 9.81 m/s^2)) * 8.4 m * cos(180°)
Work_friction = -234.12 J
Note that the negative sign is used because the force of friction opposes the motion of the crate. So, the work done by the force of friction on the crate is -234.12 J.
c) The work done by the puller on the crate can be calculated using the formula: Work = Force * Distance * cos(theta), where theta is the angle between the force and the direction of displacement. In this case, the pulling force is given as 131.0 N and it makes an angle of 12.3° with the horizontal. The distance is given as 8.4 m. Therefore, the work done by the puller can be calculated as:
Work_puller = 131.0 N * 8.4 m * cos(12.3°)
Work_puller = 1119.28 J
So, the work done by the puller on the crate is 1119.28 J.
d) The change in kinetic energy of the crate is equal to the difference in work done by all forces acting on it. Since the work done by gravity is 1152.10 J, the work done by friction is -234.12 J, and the work done by the puller is 1119.28 J, we can calculate the change in kinetic energy:
Change in KE = Work_gravity + Work_friction + Work_puller
Change in KE = 1152.10 J + (-234.12 J) + 1119.28 J
Change in KE = 2037.26 J
So, the change in kinetic energy of the crate is 2037.26 J.
e) To find the speed of the crate after it is pulled 8.4 m, we can use the principle of conservation of energy. The initial kinetic energy of the crate is (1/2) * mass * initial velocity squared, and the final kinetic energy is (1/2) * mass * final velocity squared. Since the initial velocity is given as 1.6 m/s, we can set up the following equation:
(1/2) * mass * initial velocity^2 = (1/2) * mass * final velocity^2
Simplifying and solving for the final velocity gives:
final velocity = sqrt((initial velocity^2) + (2 * Change in KE / mass))
final velocity = sqrt((1.6 m/s)^2 + (2 * 2037.26 J / 14.0 kg))
final velocity = 4.64 m/s
So, the speed of the crate after it is pulled 8.4 m is 4.64 m/s.
Fap = 131N = Applied force.
Fc=mg=14kg * 9.8N/kg=137.2N @12.3deg.
Fp = 137.2sin12.3 = 29.2N = Force parallel to the plane.
Fv = 137.2cos12.3 = 134.1N = Force
perpendicular to the plane.
Ff = u*Fv = 0.26 * 134.1 = 34.9N = Force of friction.
Fn = Fap - Fp - Ff,
Fn = 131 - 29.2 - 34.9 = 66.9N = Net force acting on crate.
a. W=Fc*h = 137.2 * 8.4sin12.3 = 246J.
b. W = Ff*d = -34.9 * 8.4 = 293J.
c. W = Fap * d = 131 * 8.4 = 1100J.
I will have to do some research on d and e. But I hope this will help.