Let the mass of the block be 7.7 kg and the angle è be 25°. Find (a) the tension in the cord and (b) the normal force acting on the block. (c) If the cord is cut, find the magnitude of the block's acceleration.

73.5

To find the tension in the cord, we can start by analyzing the forces acting on the block.

(a) The gravitational force acting on the block can be calculated using the formula Fg = m * g, where m is the mass of the block (7.7 kg) and g is the acceleration due to gravity (9.8 m/s^2). Therefore, Fg = 7.7 kg * 9.8 m/s^2 = 75.46 N.

Next, we need to consider the components of the gravitational force that are parallel and perpendicular to the inclined plane. The component of the gravitational force acting parallel to the plane is given by F_parallel = Fg * sin(è), where è is the angle (25°). Therefore, F_parallel = 75.46 N * sin(25°) = 31.94 N.

Since the block is in equilibrium (not accelerating along the inclined plane), the tension in the cord must be equal in magnitude and opposite in direction to the parallel component of the gravitational force. Therefore, the tension in the cord is 31.94 N.

(b) The normal force acting on the block is the force exerted by the inclined plane perpendicular to its surface. In this case, the normal force equals the weight of the block. Therefore, the normal force is 75.46 N.

(c) If the cord is cut, the only force acting on the block along the inclined plane is the component of the gravitational force parallel to the plane. To find the magnitude of the block's acceleration, we can use Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration (Fnet = m * a).

In this case, the net force is the parallel component of the gravitational force (31.94 N), and the mass of the block is 7.7 kg. Therefore, 31.94 N = 7.7 kg * a. Solving for 'a', we find that the magnitude of the block's acceleration is a = 31.94 N / 7.7 kg = 4.15 m/s^2.