a box of mass 7 kg is pulled up a frictionless 30 degree incline by a force of 381 N which acts parallel to the incline surface. the box has an initial speed of 5m/s and is pulled 9m. what is the speed of the box at the end of the 9m.
To find the speed of the box at the end of the 9m, we need to calculate the work done on the box and use the work-energy principle. The work done on an object is defined as the product of force applied and the displacement in the direction of the force. In this case, the force is acting parallel to the incline surface.
Step 1: Find the component of the force parallel to the incline.
The force applied, F = 381 N
The angle of the incline, θ = 30 degrees
The component of the force parallel to the incline, F_parallel = F * sin(θ)
F_parallel = 381 N * sin(30°)
F_parallel = 381 N * 0.5
F_parallel = 190.5 N
Step 2: Calculate the work done on the box.
The work done on the box, W = F_parallel * d
where d is the displacement of the box, which is given as 9 m.
W = 190.5 N * 9 m
W = 1714.5 N*m (or Joules)
Step 3: Apply the work-energy principle.
According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy.
Initial kinetic energy, K1 = 0.5 * m * v1^2
where m is the mass of the box and v1 is the initial speed of the box.
Final kinetic energy, K2 = 0.5 * m * v2^2
where v2 is the final speed of the box.
The change in kinetic energy, ΔK = K2 - K1
But K1 = 0, as the initial speed is given as 0 m/s.
ΔK = K2 - 0
ΔK = K2
Therefore, W = ΔK
1714.5 N*m = 0.5 * 7 kg * v2^2
Step 4: Solve for v2.
1714.5 N*m = 0.5 * 7 kg * v2^2
1714.5 N*m = 3.5 kg * v2^2
Divide both sides by 3.5 kg:
489 N*m/kg = v2^2
Take the square root of both sides:
22 m/s = v2
Therefore, the speed of the box at the end of the 9m is 22 m/s.
To find the final speed of the box at the end of the 9m distance, we can use the principles of work and energy.
1. First, let's calculate the work done on the box. The work done (W) is given by the formula: W = force * distance * cos(angle).
W = 381 N * 9 m * cos(30 degrees)
W = 381 N * 9 m * (sqrt(3)/2)
W ≈ 2939.26 J
2. The work done on the box is equal to the change in its kinetic energy (ΔKE). The initial kinetic energy (KEi) is given by: KEi = 1/2 * mass * initial velocity^2.
KEi = 0.5 * 7 kg * (5 m/s)^2
KEi = 87.5 J
3. The final kinetic energy (KEf) can be found by subtracting the initial kinetic energy from the work done: ΔKE = KEf - KEi.
ΔKE = 2939.26 J - 87.5 J
ΔKE ≈ 2851.76 J
4. Finally, we can find the final speed (vf) of the box using the formula for kinetic energy: KEf = 1/2 * mass * final velocity^2.
2851.76 J = 0.5 * 7 kg * vf^2
vf^2 = 2851.76 J / (0.5 * 7 kg)
vf^2 ≈ 815.03 m^2/s^2
Taking the square root of both sides, we find:
vf ≈ √(815.03 m^2/s^2)
vf ≈ 28.55 m/s
Therefore, the speed of the box at the end of the 9m distance is approximately 28.55 m/s.
weight = 7 * 9.81 Newtons
force down slope = weight * sin 30
= (1/2)weight
force up slope = 381 N
net force up slope
= 381 -(1/2) weight
= 347 N
a = F/m = 347/7 = 49.5 m/s^2
v = 5 + a t
v = 5 + 49.5 t
distance = average speed * t
9 = (1/2)(v+5) t
18 = (10+49t) t
49 t^2 + 10 t - 18 = 0
t = [ -10 +/-sqrt(100+3528) ]/98
t = [-10 + 60.2 ]/98
t = .513 second
v = 5 + 49.5(.513) = 30.4 m/s