A 2.0-kg block slides on a frictionless 15$\degree$inclined plane.A force acting parallel to the incline is applied to the block.The acceleration of the block is 1.5 m/s² down the incline.What is the applied force?

Question

Quizplus · Accepted Answer

We can use Newton's second law to solve this problem:

ΣF = ma

where ΣF is the net force acting on the block, m is the mass of the block, and a is the acceleration of the block.

We can break the weight of the block into two components: one parallel to the incline and one perpendicular to the incline:

Fg// = mg sin θ
Fg⊥ = mg cos θ

where θ is the angle of the incline.

Since the block is sliding down the incline, the applied force must be greater than the force of friction:

Fapplied > Ff = 0

Therefore, the net force is:

ΣF = Fapplied - Fg//

Substituting the given values, we have:

ΣF = Fapplied - (2.0 kg)(9.8 m/s²)(sin 15°) = (2.0 kg)(1.5 m/s²)

Solving for Fapplied, we get:

Fapplied = (2.0 kg)(1.5 m/s²) + (2.0 kg)(9.8 m/s²)(sin 15°) = 20.4 N

Since the block is moving down the incline, the applied force must be in the same direction:

Answer: C) 2.1 N up the incline.

A 20-Kg Block Slides on a Frictionless 15 °\degree° Inclined Plane

A 20-Kg Block Slides on a Frictionless 15 $\degree$ Inclined Plane