a 15-kg block is on a frictionless ramp that is inclined at 20° above the horizontal. it is connected by a very light string over an ideal pulley at the top edge of the ramp to a hanging 19-kg block, as shown in the figure. the string pulls on the 15-kg block parallel to the surface of the ramp. find the magnitude of the acceleration of the 19-kg block after the system is gently released?

Hi there!

Since the string is light and there is no friction in the pulley, the acceleration of the system is equal to the acceleration of both blocks.

We can begin by summing the forces of each block:

Block on incline:

- Force of gravity (in the negative direction away from the acceleration)

- Force of Tension

∑F = -M₁gsinФ + T

Block hanging:

- Force of gravity (Positive, in direction of acceleration)

- Force of Tension (Negative, opposite from acceleration)

∑F = M₂g - T

Sum both of these net forces for each block:

∑Fт = -M₁gsinФ + T - T + M₂g

∑Fт = -M₁gsinФ + M₂g

Divide by the mass to solve for acceleration:

[tex]a = \frac{-M_1gsin\theta+M_2g}{M_1+M_2}[/tex]

Plug in the given values:

[tex]a = \frac{-(15)(9.81)(sin20)+19(9.81)}{15+19} = 4.002 m/s^2[/tex]

The magnitude of the acceleration of the 19-kg block is 4.0 m/[tex]s^{2}[/tex].

The 15-kg block is accelerating down the ramp, and the 19-kg block is accelerating up. The string connecting the two blocks is taut, so it must be applying a force to both blocks. The force of the string on the 15-kg block is equal to the force of the string on the 19-kg block.

Let's call the force of the string on the 15-kg block T. We can use Newton's second law to write two equations for the acceleration of the two blocks:

For the 15-kg block: T - mg sin(θ) = 15a

For the 19-kg block: T = 19a

where:

m is the mass of the block

g is the acceleration due to gravity

θ is the angle of the ramp

a is the acceleration of the block

Plugging the second equation into the first equation, we get:

19a - mg sin(θ) = 15a

Solving for a, we get:

a = (mg sin(θ)) / (4m)

= (15 kg * 9.8  m/[tex]s^{2}[/tex] * sin(20°)) / (4 * 15 kg)

= 4.0 m/[tex]s^{2}[/tex]

Therefore, the magnitude of the acceleration of the 19-kg block is 4.0  m/[tex]s^{2}[/tex].

To know more about acceleration here

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