Question

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?

Answer 1

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:

$a = \frac{-M_1gsin\theta+M_2g}{M_1+M_2}$

Plug in the given values:

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