A 2.5 kg block is launched up a ramp at 9.0 m/s. If the ramp is 35 degrees above the horizontal and the coefficient of static friction is 0.60, while the coefficient of kinetic friction is 0.25.
a. How far does the block go up the ramp from its starting point?
b. What speed does it have when it slides back down to its starting point?
To answer these questions, we need to analyze the forces acting on the block.
First, let's find the force of gravity acting on the block:
F_gravity = m * g
where:
m = mass of the block = 2.5 kg
g = acceleration due to gravity = 9.8 m/s^2
F_gravity = 2.5 kg * 9.8 m/s^2
F_gravity = 24.5 N
Next, let's determine the normal force acting on the block. The normal force is equal to the force perpendicular to the ramp from the block's weight.
N = F_gravity * cos(theta)
where:
theta = angle of the ramp = 35 degrees
N = 24.5 N * cos(35 degrees)
N = 20.02 N
The force of friction depends on whether the block is at rest or in motion.
a. To find how far the block goes up the ramp from its starting point, we need to determine the maximum force of static friction between the block and the ramp. The maximum force of static friction is given by:
F_static_friction_max = coefficient of static friction * N
where the coefficient of static friction = 0.60
F_static_friction_max = 0.60 * 20.02 N
F_static_friction_max = 12.01 N
The force applied to move the block up the ramp is equal to the component of the force of gravity parallel to the ramp. Let's calculate this force:
F_parallel = F_gravity * sin(theta)
F_parallel = 24.5 N * sin(35 degrees)
F_parallel = 13.95 N
Since the force to move the block up the ramp is less than the maximum force of static friction, the block stays at rest. Therefore, the block doesn't move up the ramp from its starting point.
b. To find the speed of the block when it slides back down to its starting point, we need to consider the force of kinetic friction. The force of kinetic friction is given by:
F_kinetic_friction = coefficient of kinetic friction * N
where the coefficient of kinetic friction = 0.25
F_kinetic_friction = 0.25 * 20.02 N
F_kinetic_friction = 5.01 N
The net force acting on the block is:
Net_force = F_gravity * sin(theta) - F_kinetic_friction
Net_force = 24.5 N * sin(35 degrees) - 5.01 N
Net_force = 8.90 N
Using Newton's second law (F = m * a), we can find the acceleration of the block:
Net_force = m * a
a = Net_force / m
a = 8.90 N / 2.5 kg
a = 3.56 m/s^2
Now, to find the speed of the block when it slides back down to its starting point, we can use the kinematic equation:
v^2 = u^2 + 2 * a * s
where:
v = final velocity (unknown)
u = initial velocity (0 m/s, since the block starts from rest)
a = acceleration = -3.56 m/s^2 (negative because the block is moving in the opposite direction)
s = distance traveled (unknown)
Plugging in the values:
v^2 = 0^2 + 2 * (-3.56 m/s^2) * s
v^2 = -7.12 * s
Since the block ends up at the same height as its starting point, the distance traveled up the ramp is equal to the distance traveled down the ramp. So, we can write the equation as:
v^2 = -7.12 * (2 * s)
Plugging in the values:
v^2 = -14.24 * s
Since the speed is positive, we take the square root of both sides:
v = sqrt(-14.24 * s)
We don't have the value of s to find the exact speed, but we can determine that as the block slides back down, its speed will depend on the distance traveled up the ramp (which we found to be 0). So, its speed when it slides back down to its starting point will be 0 m/s.