A 10 kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20 degrees with the horizontal. If the coefficient of kinetic friction is .40 and the crate is pulled a distance of 5 m, (a) how much work is done by gravitational force? (b) How much work is done by the 100 N force? (c) What is the change in kinetic energy of the crate? (d) What is the speed of the crate after it is pulled 5m?
friction force down the plane:
mu*mg*Cos20
gravity force down the plane:
mg*sin20
work done by 100N force:
100*5=500J
work=force*distance
a. work done by gravity:
mg*sin20*(-5)=-5mg*sin20
b. work done by 100N force:
500J
c.KE change:
500J-5mg*sin20-mu*mg*cosTheta*5
d. speed final
1/2 m v^2=KEchange
solve for v
To solve this problem, we will use the work-energy theorem. The work done by a force is equal to the change in kinetic energy of an object.
(a) To calculate the work done by the gravitational force, we need to determine the change in potential energy. The potential energy of an object is given by the formula: potential energy = mass × gravity × height. In this case, the height is zero because the crate is pulled horizontally. Therefore, the work done by the gravitational force is zero.
(b) The work done by the 100 N force can be found using the formula: work = force × distance × cos(angle). In this case, the force is 100 N, and the angle is 20 degrees. The distance is given as 5 m. So, plug in the values into the formula: work = 100 N × 5 m × cos(20 degrees). Calculate using a calculator to find the work done by the 100 N force.
(c) To find the change in kinetic energy, we need to subtract the initial kinetic energy from the final kinetic energy. The initial kinetic energy can be found using the formula: kinetic energy = 0.5 × mass × speed^2. The mass of the crate is 10 kg, and the initial speed is given as 1.5 m/s. Plug in these values into the formula to calculate the initial kinetic energy.
To find the final kinetic energy, we can use the work-energy theorem. The work done by the friction force is equal to the change in kinetic energy. The work done by the friction force can be calculated using the formula: work = force of friction × distance. The force of friction can be found by multiplying the coefficient of kinetic friction (0.40) by the normal force (which is equal to the weight of the crate, which is mass × gravity).
To calculate the final speed, we can use the formula: final speed^2 = initial speed^2 + 2 × acceleration × distance. The acceleration can be found using the formula: acceleration = net force / mass. The net force is the difference between the force parallel to the incline and the force of friction.
(d) Plug in the values into the formulas and calculate the change in kinetic energy, as well as the final speed using a calculator.
To answer these questions, we need to break down the problem into smaller steps and use some physics principles. Let's start with the first question:
(a) How much work is done by gravitational force?
The work done by the gravitational force can be calculated using the formula: work = force * distance * cos(angle).
Given:
- Mass of the crate (m) = 10 kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Distance (d) = 5 m
- Angle (θ) = 0 degrees (because we are only considering the force parallel to the incline)
The gravitational force acting on the crate can be calculated using the formula: force_gravity = mass * gravity.
So, force_gravity = 10 kg * 9.8 m/s² = 98 N.
Now, we can calculate the work done by gravity:
work_gravity = force_gravity * distance * cos(θ) = 98 N * 5 m * cos(0) = 490 J.
Therefore, the work done by the gravitational force is 490 Joules.
(b) How much work is done by the 100 N force?
The work done by a force can be calculated using the formula: work = force * distance * cos(angle).
Given:
- Force (F) = 100 N
- Distance (d) = 5 m
- Angle (θ) = 20 degrees
Now, let's calculate the work done by the 100 N force:
work_force = force * distance * cos(angle) = 100 N * 5 m * cos(20 degrees).
To calculate cos(20 degrees), you can either use a scientific calculator or a trigonometric identity table. Using a calculator, cos(20 degrees) is approximately 0.9397.
So, work_force = 100 N * 5 m * 0.9397 ≈ 469.85 J.
Therefore, the work done by the 100 N force is approximately 469.85 Joules.
(c) What is the change in kinetic energy of the crate?
The change in kinetic energy can be calculated using the formula: ΔKE = work_net.
The net work done on an object is equal to the sum of all works done on it.
So, net work = work_force + work_gravity.
Substituting the known values:
net work = 469.85 J + 490 J = 959.85 J.
Therefore, the change in kinetic energy of the crate is 959.85 Joules.
(d) What is the speed of the crate after it is pulled 5 m?
To calculate the final speed of the crate, we can use the work-energy principle.
The work done on an object is equal to the change in its kinetic energy (ΔKE). Therefore, using the work-energy principle:
work_net = ΔKE.
Substituting the known values:
959.85 J = ΔKE.
The initial kinetic energy of the crate is given as 1/2 * mass * initial velocity^2.
Given:
- Mass of the crate (m) = 10 kg
- Initial velocity (v_initial) = 1.5 m/s
Initial kinetic energy (KE_initial) = 1/2 * 10 kg * (1.5 m/s)^2 = 11.25 J.
So, the final kinetic energy (KE_final) = KE_initial + ΔKE = 11.25 J + 959.85 J = 971.1 J.
Since KE_final = 1/2 * mass * final velocity^2, we can rearrange the equation to find the final velocity (v_final):
v_final = √(2 * KE_final / mass).
Calculating:
v_final = √(2 * 971.1 J / 10 kg) ≈ 13.97 m/s.
Therefore, the speed of the crate after it is pulled 5 m is approximately 13.97 m/s.