A 15 kg block is on a ramp which is inclined at 20 degrees above the horizontal. It's connected by a string to a 20kg mass which hangs over the top edge of the ramp. With no frictional forces taken into account, what is the magnitude of the acceleration of the 20kg block?

Question

A 15 kg block is on a ramp which is inclined at 20 degrees above the horizontal. It's connected by a string to a 20kg mass which hangs over the top edge of the ramp. With no frictional forces taken into account, what is the magnitude of the acceleration of the 20kg block?

Answer 1

To find the magnitude of the acceleration of the 20 kg block, we need to consider the forces acting on it.

First, let's draw a free-body diagram for the 20 kg block.

1. Gravitational force (weight): The weight of the block is given by the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s²). In this case, the weight is W = 20 kg × 9.8 m/s².

2. Tension force: The tension in the string connecting the two blocks is the same throughout the string. So, the 20 kg block will experience the same tension force as the 15 kg block.

Now, let's consider the forces acting along the ramp.

1. Normal force: The normal force is the force exerted by the ramp perpendicular to the surface. Since there is no vertical acceleration, the normal force cancels out the vertical component of weight. Therefore, the normal force is equal to the vertical component of the weight of the 15 kg block, which is given by N = mg × cos(θ), where θ is the angle of the incline.

2. Frictional force: As mentioned in the problem statement, frictional forces are not taken into account. So, we can assume there is no friction in this case.

Now, let's resolve the weight of the 20 kg block into components.

The vertical component of weight is given by W₁ = 20 kg × 9.8 m/s² × sin(θ).
The horizontal component of weight is given by W₂ = 20 kg × 9.8 m/s² × cos(θ).

Since there is no friction, the net force in the horizontal direction is zero.
So, the horizontal component of the tension force must be equal to the horizontal component of the weight:

T cos(θ) = W₂

Now, we can solve for T:

T = W₂ / cos(θ)
= (20 kg × 9.8 m/s² × cos(θ)) / cos(θ)
= 20 kg × 9.8 m/s²

Since the 20 kg block is connected to the 15 kg block, the tension force is the force accelerating the system. The magnitude of the acceleration is given by the formula:

acceleration = T / (15 kg + 20 kg)
= (20 kg × 9.8 m/s²) / (15 kg + 20 kg)

Now, calculate the magnitude of the acceleration.