A block of mass M is projected up a frictionless inclined plane with a speed VO. The angle of incline is given as theta.
a. How far up the plane does it go?
b. How much time does it take to get there?
I was stuck after Fg(cos(theta+90)= Ma_x
gcos(90+theta)/M = a_x
can someone please help me
kinetic energy at bottom = (1/2) M Vo^2
= potential energy at to when it stops
= m g h = M g (x sin theta)
so
(1/2) M Vo^2 = M g x sin thets
x = Vo^2/(2 g sin theta)
average speed = Vo/2
time = distance/average speed
= Vo^2/(2 g sin theta)] / [ Vo/2)]
= Vo/(g sin theta)
thanks but my instrutor don't want me to use energy to solve the problem since we havnt learn it yet, can you do it using newton's law?
Certainly! Let's break down the problem step by step to find the answers to both parts.
a. How far up the plane does the block go?
To determine the distance the block travels up the inclined plane, we can use the equations of motion. Since there is no friction, the only force acting on the block is its weight component parallel to the inclined plane, which we can call Fg_parallel.
To find Fg_parallel, we need to resolve the weight (Fg) of the block into its components. The weight of the block is given by Fg = M * g, where M is the mass of the block and g is the acceleration due to gravity.
Now, let's resolve the weight vector into its components. The weight vector can be broken down into two components: Fg_parallel, acting parallel to the inclined plane, and Fg_perpendicular, acting perpendicular to the inclined plane. We get:
Fg_parallel = Fg * sin(theta)
Fg_perpendicular = Fg * cos(theta)
Since the inclined plane is frictionless, there is no force parallel to the plane acting on the block, except the component of weight parallel to the plane (Fg_parallel).
The net force acting on the block in the direction parallel to the inclined plane is equal to the product of its mass (M) and acceleration (a). Therefore, we have:
Fg_parallel = M * a
Substituting the values of Fg_parallel and Fg into the above equation, we get:
M * g * sin(theta) = M * a
Canceling out the mass from both sides, we get:
g * sin(theta) = a
The acceleration (a) can also be expressed as the derivative of velocity (v) with respect to time (t):
a = dv/dt
Integrating both sides of the equation, we can solve for the velocity (v) as a function of time (t):
v = ∫(g * sin(theta)) dt
v = g * sin(theta) * t
Now, let's find the time it takes for the block to reach the highest point (t_up) where its velocity becomes zero. At this point, the block will momentarily stop before reversing its direction due to gravity.
When the block reaches the highest point, its final velocity (vf) is zero. We can use the kinematic equation to find the time taken to reach this point:
vf = vo + a * t_up
Since vf = 0 at the highest point, the equation becomes:
0 = vo + (-g * sin(theta) * t_up)
Simplifying, we have:
t_up = vo / (g * sin(theta))
Now, to find the distance traveled up the inclined plane, we can use the equation of motion:
s = vo * t_up + 1/2 * (-g * sin(theta)) * t_up^2
Substituting the value of t_up into the equation, we get:
s = vo * (vo / (g * sin(theta))) + 1/2 * (-g * sin(theta)) * (vo / (g * sin(theta)))^2
Simplifying further, we have:
s = vo^2 / (2g * sin(theta))
Therefore, the distance the block travels up the inclined plane is given by:
s = vo^2 / (2g * sin(theta))
b. How much time does it take to get there?
As we calculated earlier, the time taken to reach the highest point (t_up) is given by:
t_up = vo / (g * sin(theta))
This is the same as the time it takes for the block to travel up the inclined plane.
To summarize:
a. The distance the block travels up the plane (s) is given by s = vo^2 / (2g * sin(theta))
b. The time it takes for the block to reach the highest point (t_up) is given by t_up = vo / (g * sin(theta))
I hope this explanation helps you understand how to approach and solve the problem.