A boy sliding down a frictionless plane inclined at an angles 35° with the horizontal. What is the acceleration of the book?
force of gravity down the slide:
mg*SinTheta
Forcedowntheslide=mass*acceleration
solve for acceleration.
If you had friction, the force of friction would be mu*mg*CosTheta, and that would subtract from the gravity force down the slide. Frictionless implies mu=0
To determine the acceleration of the boy sliding down the inclined plane, we need to break down the forces acting on the boy.
First, we need to resolve the weight of the boy into components along the incline and perpendicular to the incline. The weight is given by the formula W = mg, where m is the mass of the boy and g is the acceleration due to gravity (9.8 m/s^2).
The component of the weight acting along the incline can be found using the formula W_parallel = W * sin(theta), where theta is the angle of the incline (35°). The component of the weight perpendicular to the incline is given by W_perpendicular = W * cos(theta).
Since the plane is frictionless, there is no force opposing the motion along the incline (horizontal) direction. Therefore, the only force acting on the boy along the incline is the component of the weight acting along the incline.
The net force along the incline (horizontal) direction is equal to the mass of the boy multiplied by the acceleration along the incline (a). According to Newton's second law, F_net = m * a.
Therefore, we can equate the component of the weight acting along the incline (W_parallel) to the net force along the incline (F_net):
W_parallel = F_net
By substituting the values:
m * g * sin(theta) = m * a
We can cancel out the mass (m) from both sides of the equation:
g * sin(theta) = a
Plugging in the known values:
g = 9.8 m/s^2 (acceleration due to gravity)
theta = 35°
a = 9.8 m/s^2 * sin(35°)
By calculating this expression, we find that the acceleration of the boy sliding down the frictionless inclined plane is approximately 5.61 m/s^2.