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 determine the speed of the box at the end of the 9m, we need to consider the forces acting on the box, including gravity and the applied force.
1. Find the component of the applied force parallel to the incline:
The applied force is acting parallel to the incline surface, so we need to find the parallel component.
F_parallel = F_applied * sin(θ), where θ is the angle of the incline (30 degrees).
F_parallel = 381 N * sin(30°)
2. Calculate the net force along the incline:
The net force is the sum of the parallel component of the applied force and the force component due to gravity.
F_net = F_parallel - F_gravity
F_gravity = m * g, where m is the mass of the box (7 kg) and g is the acceleration due to gravity (9.8 m/s^2).
3. Calculate the work done:
Work done = F_net * distance, where distance is 9m.
4. Calculate the change in kinetic energy:
The work done is equal to the change in kinetic energy.
ΔKE = Work done.
5. Calculate the final speed of the box:
Using the formula for kinetic energy, KE = 1/2 * m * v^2, where v is the final speed.
The change in kinetic energy can be written as ΔKE = KE_final - KE_initial.
Now, we can solve the equations to find the final speed of the box:
ΔKE = KE_final - KE_initial
Work done = ΔKE
F_net * distance = 1/2 * m * v^2 - 1/2 * m * (5 m/s)^2
(F_parallel - F_gravity) * distance = 1/2 * m * v^2 - 1/2 * m * (5 m/s)^2
Substituting the values we know:
(F_parallel - F_gravity) * distance = 1/2 * 7 kg * v^2 - 1/2 * 7 kg * (5 m/s)^2
Now, plug in the values and solve for v.
Initial KE + Work added = final KE + GPE
1/2(7)5^2 + 381*9 = 1/2(7)vf^2 + 7(9.8)h
Find h using trig for a triangle with 30o angle and 9 m hypotenuse.