(Q1)A 40N force applied at an angle of 37c above the horizontal pulls a 5kg box on a horizontal floor. The acceleration of the box is 3m/s2,How large a frctional force must be retarding the motion of the box. (a)50 N(b)13N(c)17N(d)25N (Q2) A block of mass 2kg is connected to a freely hanging block of mass 4kg by a light and inextensible string which passes over pulley at the edge of a table. The 2kg mass is on the surface of the table assumed to be smootm.Calculate the acceleration of the system and the tension in the string (a)6.7m/s2 and13.3N(b)3.3 m/s2 and 34.4N (c)0.54m/s2 40.6N(d)2.5 m/s2 and 32.2N (Q3)A 20kg block on an inclined plane is pulled up the plane with a rope tied to the block. The rope is at angle of 37c above the surface of the plane.The tension in the rope is 250N and the frictional force on the block is 8.0N.What is the acceleration of the block?
1. Fn = m*a
40*cos37-Fk = 5 * 3
31.9-Fk = 15
Fk = 31.9-15 = 17 N.
3. a = Fn/m
a = (250*cos37-8)/20 = 9.58 m/s^2.
To solve these questions, we'll need to apply Newton's laws of motion and some basic principles in physics. Let's start with each question one by one.
(Q1) To find the fractional force retarding the motion of the box, we first need to calculate the total force acting on the box and then subtract the force responsible for acceleration.
The force acting on the box can be divided into two components: the horizontal component and the vertical component. The horizontal component is given by F_horizontal = F * cos(37°), where F is the applied force (40 N). Substituting the values, F_horizontal = 40 N * cos(37°) = 40 N * 0.7986 ≈ 31.95 N (rounded to two decimal places).
The vertical component doesn't affect the horizontal motion and can be ignored in this case.
Now, the frictional force retarding the motion of the box is equal and opposite to the horizontal component of the applied force. Therefore, the frictional force is equal to F_friction = -F_horizontal = -31.95 N (negative sign indicates opposite direction).
Hence, the answer is (b) 13 N.
(Q2) In this case, we have a system of two masses connected by a string over a pulley. When solving problems like these, it's helpful to consider the forces acting on each mass individually.
For the 2 kg mass on the table, the only force acting on it is the tension in the string (Tension_1). Using Newton's second law (F = ma), we have Tension_1 = m_1 * a, where m_1 is the mass (2 kg) and a is the acceleration.
For the 4 kg mass hanging freely, we have the force due to its weight (m_2 * g) acting downward and the tension in the string (Tension_2) acting upward. Using Newton's second law, we have Tension_2 - m_2 * g = m_2 * a, where m_2 is the mass (4 kg), g is the acceleration due to gravity (9.8 m/s^2), and a is the acceleration.
Since the masses are connected by the same string, the tension in the string is the same for both masses (Tension_1 = Tension_2 = Tension). Therefore, we can equate the two equations above to find the acceleration.
m_1 * a = m_2 * a + m_2 * g
2 kg * a = 4 kg * a + 4 kg * 9.8 m/s^2
2a - 4a = 39.2 m/s^2
-2a = -39.2 m/s^2
a = 19.6 m/s^2 (divided by -2)
Now, we can substitute the acceleration into either equation to find the tension in the string.
Tension = Tension_1 = m_1 * a
Tension = 2 kg * 19.6 m/s^2
Tension = 39.2 N
Hence, the answer is (d) 2.5 m/s^2 and 32.2 N.
(Q3) In this case, we need to consider the forces acting on the block on the inclined plane. The forces at play are the gravitational force (m * g) acting straight downward, the tension in the rope (Tension) acting upward, and the frictional force (F_friction) acting opposite to the motion of the block.
We can start by resolving the gravitational force into components. The component parallel to the inclined plane is m * g * sin(37°), and the component perpendicular to the inclined plane is m * g * cos(37°). The tension in the rope (Tension) also has components: Tension_parallel = Tension * cos(37°) and Tension_perpendicular = Tension * sin(37°). The frictional force (F_friction) is given as 8.0 N.
Now, let's apply Newton's second law in the direction of motion. The net force F_net is given by the difference between the parallel component of the tension and the frictional force (F_net = Tension_parallel - F_friction). This force is responsible for the acceleration of the block.
F_net = Tension_parallel - F_friction
F_net = Tension * cos(37°) - 8.0 N
Now, we can relate the net force to the acceleration of the block using Newton's second law (F_net = m * a), where m is the mass (20 kg). Solving for the acceleration:
F_net = m * a
Tension * cos(37°) - 8.0 N = 20 kg * a
Now we have one equation and one unknown (acceleration). To solve for acceleration, we need to know the value of tension (Tension). However, tension is not given in the question, so we cannot determine the exact value of the acceleration.
In conclusion, without knowing the exact value of the tension, we cannot calculate the acceleration of the block.