A horizontal force of 99 N pushes a 11 kg block up a frictionless incline that makes an angle of 51° with the horizontal.
(a) What is the normal force that the incline exerts on the block?
(b) What is the acceleration of the block? (Hint: assume up the block as the positive direction.)
To find the answers to these questions, we can use principles from Newton's laws of motion and trigonometry. Let's break down the steps to get the answers.
(a) To find the normal force exerted by the incline on the block, we need to consider the forces acting on the block. In this case, there are two forces at play: the gravitational force mg (where m is the mass and g is the acceleration due to gravity) and the force applied horizontally (99 N).
Since the incline is frictionless, there is no horizontal force component to consider. Therefore, the normal force (N) and gravitational force (mg) must balance each other in the vertical direction. Considering the given angle of 51°, we can use the trigonometric relationship:
cos(51°) = N/mg
Rearranging the equation, we get:
N = mg * cos(51°)
Substituting the given values, we have:
mass (m) = 11 kg
acceleration due to gravity (g) = 9.8 m/s^2
angle (θ) = 51°
N = (11 kg) * (9.8 m/s^2) * cos(51°)
Calculating this expression will give us the value for the normal force.
(b) To find the acceleration of the block, 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.
In this case, the net force acting on the block is the horizontal force exerted (99 N) minus the force component due to gravity acting down the inclined plane (mg * sin(θ)). So the equation becomes:
net force = ma
99 N - mg * sin(θ) = ma
Substituting the given values:
mass (m) = 11 kg
acceleration due to gravity (g) = 9.8 m/s^2
angle (θ) = 51°
We can then solve for the acceleration (a) using this equation.
By following these steps, you should be able to calculate the values for both the normal force and the acceleration of the block.