A block is pushed up a frictionless incline by

an applied horizontal force as shown.
The acceleration of gravity is 9.8 m/s2 .
mass=2.7 kg
F = 33 N
theta=29◦
What is the magnitude of the resulting ac-
celeration of the block?
Answer in units of m/s2.

To find the magnitude of the resulting acceleration of the block, we need to 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.

The force acting on the block along the incline can be resolved into two components: one parallel to the incline's surface and one perpendicular to it.

The perpendicular component of the applied force will cancel out the perpendicular component of the gravitational force, as the block is on a frictionless incline. Therefore, the perpendicular components will not contribute to the acceleration.

The parallel component of the applied force, F_parallel, will accelerate the block up the incline. To find this component, we can use a trigonometric relationship. The parallel component is given by:

F_parallel = F * sin(theta)

where F is the applied force and theta is the angle of the incline.

Now, we can substitute the given values into the equation:

F_parallel = 33 N * sin(29°) ≈ 16.22 N

Next, we can use Newton's second law to find the acceleration. The net force acting on the block is the parallel component of the applied force:

F_net = F_parallel

Since the net force is equal to the mass of the block multiplied by its acceleration, we have:

F_net = m * a

Substituting the given mass:

16.22 N = 2.7 kg * a

Finally, we can solve for the acceleration:

a = 16.22 N / 2.7 kg ≈ 6 m/s^2

Therefore, the magnitude of the resulting acceleration of the block is approximately 6 m/s^2.