a 20 kg traveler's bag is being pulled along the floor with a force of 50 NF. the force is applied on the handle forming an angle of 30 degree with the horizontal. neglecting friction,what is the acceleration og the bag? How much force is exerted by the floor on the bag?
M*g = 20 * 9.8 = 196 N.=Wt. of suitcase.
a. a = Fx/M = 50*cos30/20 = 2.17 m/s^2.
b. Fn = 196 - 50*sin30 = 171 N.=Normal
force = Force exerted on the bag by the
floor.
NONE
2.17 m/s^2
To find the acceleration of the bag, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
1. First, we need to resolve the force applied on the bag into horizontal and vertical components. The horizontal component can be found using trigonometry:
F_horizontal = F_applied * cos(angle)
= 50 N * cos(30°)
≈ 50 N * 0.866
≈ 43.3 N
The vertical component can be found similarly:
F_vertical = F_applied * sin(angle)
= 50 N * sin(30°)
≈ 50 N * 0.5
= 25 N
2. Since there is no friction, the horizontal force is the only force acting in the direction of motion. So, we can equate this force to the mass of the bag (20 kg) multiplied by its acceleration:
F_horizontal = m * a
43.3 N = 20 kg * a
Solving for acceleration:
a = 43.3 N / 20 kg
a ≈ 2.165 m/s²
Therefore, the acceleration of the bag is approximately 2.165 m/s².
3. The force exerted by the floor on the bag can be found by considering the vertical forces. In this case, the force exerted by the floor on the bag is equal in magnitude and opposite in direction to the vertical component of the applied force. So, the force exerted by the floor is 25 N.