A small block with mass 0.308 kg is sliding down a frictionless ramp that is inclined at an angle of 53.1° above the horizontal. Assume
g = 9.80 m/s2.
(a) As the object slides down the incline, what is the magnitude of the normal force that the surface of the incline exerts on the block?
(b) As the object slides down the incline, what is the magnitude of the component of the gravity force parallel to the surface of the incline?
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To find the magnitude of the normal force exerted by the surface of the incline on the block:
Step 1: Draw a diagram of the situation. Label the angles and forces involved:
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Step 2: Identify the forces acting on the block. In this case, we have two forces: the weight of the block (mg) pointing downward and the normal force (N) exerted by the surface of the incline.
Step 3: Set up the equations using the components of the gravitational force and resolve them parallel and perpendicular to the incline.
Perpendicular to the incline (normal direction):
N - mg*cos(theta) = 0
Parallel to the incline:
mg*sin(theta) = 0
where theta is the angle of inclination, which is given as 53.1°.
Step 4: Solve the equations for the unknowns.
Rearrange the first equation:
N = mg*cos(theta)
Substitute the given values:
N = (0.308 kg) * (9.80 m/s^2) * cos(53.1°)
N = 1.595 N
Therefore, the magnitude of the normal force is 1.595 N.
To find the magnitude of the component of the gravitational force parallel to the surface of the incline:
Step 1: Use the same diagram from before.
Step 2: Identify the forces acting on the block. In this case, we have two forces: the weight of the block (mg) pointing downward and the normal force (N) exerted by the surface of the incline.
Step 3: Resolve the weight vector into its components. The component parallel to the incline will be mg*sin(theta), where theta is the angle of inclination.
Step 4: Solve for the unknown.
Substitute the given values:
mg*sin(theta) = (0.308 kg) * (9.80 m/s^2) * sin(53.1°)
mg*sin(theta) = 1.963 N
Therefore, the magnitude of the component of the gravity force parallel to the surface of the incline is 1.963 N.
To solve this problem, we will use the concepts of forces and the components of forces acting on an inclined plane.
(a) The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is perpendicular to the surface of the incline. To find the magnitude of the normal force, we need to find the component of the gravitational force perpendicular to the incline.
The gravitational force can be calculated using the formula:
F = m * g,
where F is the gravitational force, m is the mass of the object, and g is the acceleration due to gravity.
Since the object is on an incline, we need to find the component of the gravitational force perpendicular to the incline. This component is given by:
F_perpendicular = F * cos(theta),
where theta is the angle of the incline.
Substituting the values, we have:
m = 0.308 kg and theta = 53.1°.
F = m * g = 0.308 kg * 9.80 m/s^2.
Now we can calculate the magnitude of the normal force:
F_perpendicular = F * cos(theta).
(b) The component of the gravitational force parallel to the surface of the incline is given by:
F_parallel = F * sin(theta).
Substituting the values, we have:
F_parallel = F * sin(theta).
Now we can calculate the magnitude of the component of the gravity force parallel to the surface of the incline.
a) mg*cosTheta
b) mg*sinTheta