A block is projected up a frictionless inclined plane with initial speed v1 = 3.40 m/s. The angle of incline is è = 31.1°.
(a) How far up the plane does it go?(m)
(b) How long does it take to get there? (s)
(c) What is its speed when it gets back to the bottom?(m/s)
its KE will equal the change in PE
1/2 m 3.4^2=mg*d/sin31.1
solve for d
How long? its average velocity is 3.4/2, the distance above, so t=distance/avgvelocity
when it returns to bottom, same speed as started, opposite direction.
To solve this problem, we can use the principles of physics, such as conservation of energy and kinematics equations. Let's break down each part of the problem one by one:
(a) How far up the plane does it go?
To find how far up the plane the block goes, we need to analyze the vertical motion. The block will eventually reach its maximum height where its final velocity becomes zero. At that point, all the initial kinetic energy will be converted into gravitational potential energy.
1. Determine the initial vertical velocity (v1y):
v1y = v1 * sin(è)
v1y = 3.40 m/s * sin(31.1°)
2. Find the time it takes (t) for the block to reach its maximum height:
Use the kinematic equation: vy = v1y + g*t, where vy is the final vertical velocity, v1y is the initial vertical velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
vy = 0 m/s (since the block reaches its maximum height, its vertical velocity becomes zero)
0 = v1y + g*t
t = -v1y / g
3. Calculate the maximum height (h) using the kinematic equation for vertical displacement:
h = v1y * t + (1/2) * g * t^2
(b) How long does it take to get there?
We have already calculated the time (t) it takes for the block to reach its maximum height. This value can be directly used as the answer.
(c) What is its speed when it gets back to the bottom?
To find the speed of the block when it gets back to the bottom, we consider the conservation of mechanical energy. The total mechanical energy at the top of the incline (where the block starts) is equal to the total mechanical energy at the bottom of the incline (where the block ends).
1. Calculate the initial kinetic energy (KE) at the top of the incline:
KE_top = (1/2) * m * v1^2, where m is the mass of the block
2. Calculate the final kinetic energy (KE) at the bottom of the incline:
KE_bottom = (1/2) * m * v2^2, where v2 is the final velocity at the bottom
3. We assume the block loses no energy due to friction or other factors. Thus, KE_top = KE_bottom.
(1/2) * m * v1^2 = (1/2) * m * v2^2
Simplifying, we can solve for v2:
v2 = v1 * sqrt(2)
Now, let's substitute the given values into the equations to find the answers:
- Given values:
v1 = 3.40 m/s
è = 31.1°
1. Calculate v1y:
v1y = 3.40 m/s * sin(31.1°)
2. Calculate t:
t = -v1y / g
3. Calculate h:
h = v1y * t + (1/2) * g * t^2
4. Calculate v2:
v2 = v1 * sqrt(2)
These calculations will give us the answers to the questions (a), (b), and (c) in meters (m) and seconds (s).