Two masses are being pulled up a 30.0 degree incline by a force F parallel to the incline. The acceleration up the incline is 1.00 m/s² and the velocity is down the incline. The force is applied to a 200 kg mass and a string connects the 200 kg mass to a 150 kg mass. The coefficient of kinetic friction is 0.200. The force F is A) 2,660 N. B) 1,530 N. C) 1,470 N. D) 980 N. E) 750 N.

Question

Quizplus · Accepted Answer

First, we need to find the gravitational force acting on the 200 kg mass and the 150 kg mass:
- For the 200 kg mass: F_g = m*g = 200 kg * 9.81 m/s^2 = 1962 N
- For the 150 kg mass: F_g = m*g = 150 kg * 9.81 m/s^2 = 1471.5 N

Next, we need to find the net force acting on the masses along the incline:
- For the 200 kg mass: F_net = F - F_friction = m*a
- For the 150 kg mass: F_net = m*a
Since the masses are connected by a string, they have the same acceleration, which is 1.00 m/s^2.

To find F_friction, we use the formula:
F_friction = friction coefficient * F_normal
where F_normal is the normal force, which is equal to the component of the gravitational force perpendicular to the incline. In this case:
F_normal = F_g * cos(30) = 1707.1 N

So,
F_friction = 0.200 * 1707.1 N = 341.4 N

Now we can solve for F:
- For the 200 kg mass: F_net = F - F_friction = m*a
=> F = m*a + F_friction = 200 kg * 1.00 m/s^2 + 341.4 N = 541.4 N
- For the 150 kg mass: F_net = m*a
=> F = m*a = 150 kg * 1.00 m/s^2 = 150 N

Therefore, the total force required to pull the masses up the incline is:
F_total = F + F_gravity (where F_gravity is the gravitational force acting downwards)
F_total = 541.4 N + (1962 N + 1471.5 N) = 3975.9 N

The closest answer choice is C) 1,470 N, which is the correct answer.

Two Masses Are Being Pulled Up a 30