Video Transcript
An object has a mass 𝑚 is equal to
1.3 kilograms. The object is on a smooth slope
with an angle 𝜃 equals 33 degrees, as shown in the diagram. What is the acceleration 𝑎 of the
object parallel to the surface of the slope?
Let’s begin to solve the problem by
drawing in the forces on our object. The weight of the object pulls
straight down, and the normal reaction force acts perpendicularly to the
surface. Since the slope is smooth, we do
not need to worry about friction. The question asks us for the
acceleration of the object parallel to the surface of the slope. Therefore, we need to look at any
forces that also happen to be parallel. Only the weight can have a
component that is parallel as a normal reaction force is perpendicular to the
surface. Let’s break down the weight into
its components. The only component that we care
about is the parallel component of the weight.
Looking at the right triangle that
we drew between the weight and its components, we can see that the parallel
component of the weight is the opposite side of the triangle to the angle 𝜃 and the
weight 𝑊 is the hypotenuse. We can use trigonometry to
determine the parallel component of the weight using sin of 𝜃 is equal to the
opposite side of the triangle divided by the hypotenuse. Plugging in our variables, we’d
have sin 𝜃 is equal to the parallel component divided by the weight. To isolate the parallel component,
we multiply both sides of the equation by 𝑊. This cancels out the 𝑊 on the
right-hand side. Our expression for the parallel
component of the weight becomes 𝑊 sin 𝜃.
We should remember that the weight
of an object is equal to the mass of the object times the acceleration due to
gravity, so we can replace the weight with 𝑚𝑔. Now we can plug in the values from
our problem. We replace mass with 1.3 kilograms,
acceleration due to gravity with 9.8 meters per second squared, and the sin of the
angle 𝜃 with the sin of 33 degrees, which gives us a value for the parallel
component of our weight of 6.94 newtons. How does that help us find the
acceleration in the parallel direction?
We should recall that Newton’s
second law is a relationship between the net force and the acceleration of an
object, 𝐹 net equals 𝑚𝑎. The net force parallel to the
surface of the slope, we just found to be 6.94 newtons. And the mass of the object was
given to us as 1.3 kilograms. To isolate the acceleration, we
divide both sides of the equation by 1.3 kilograms, thereby canceling out 1.3
kilograms on the right-hand side. When we divide our two numbers, we
get 5.3 meters per second squared. The acceleration of the object
parallel to the surface of the slope is 5.3 meters per second squared.
