A 6 kg black is placed at the very top of a ramp that is inclined at 31 degrees and the block is released to slide down the ramp. As the block slides friction exerts a constant resistive force of 17N directed back up the ramp
What is the block's speed when it arrives at the bottom of the 3.5 meter long ramp?
the gravitational potential energy at the top of the ramp becomes kinetic energy at the bottom of the ramp (minus the work done by friction)
m g h - f d = 1/2 m v^2
[6 * 9.8 * 3.5 sin(31º)] - (17 * 35.) = 1/2 * 6 * v^2
To find the block's speed when it arrives at the bottom of the ramp, we can use the principles of physics.
Step 1: Resolve the weight of the block
Given:
- Mass of the block (m) = 6 kg
- Acceleration due to gravity (g) = 9.8 m/s^2
The weight of the block (W) can be calculated using the formula:
W = m * g
Substituting the given values, we get:
W = 6 kg * 9.8 m/s^2
W = 58.8 N
Step 2: Resolve the weight components
The weight of the block has two components due to the inclined ramp - one perpendicular to the ramp (normal force) and the other parallel to the ramp (force of gravity).
The component of the weight parallel to the ramp (W_parallel) can be calculated using the formula:
W_parallel = W * sin(θ)
Given:
- Angle of the ramp (θ) = 31 degrees
Substituting the known values, we get:
W_parallel = 58.8 N * sin(31 degrees)
W_parallel ≈ 30.28 N
Step 3: Determine the net force acting on the block
The net force acting on the block is the difference between the force parallel to the ramp and the resistive force (friction) acting in the opposite direction.
Net force (F_net) = W_parallel - F_friction
Given:
- Resistive force (friction) = 17 N
Substituting the known values, we get:
F_net = 30.28 N - 17 N
F_net ≈ 13.28 N
Step 4: Apply Newton's second law of motion
The net force acting on the block is responsible for its acceleration. Using Newton's second law, we can relate force, mass, and acceleration.
F_net = m * a
Rearranging the formula, we can solve for acceleration:
a = F_net / m
Substituting the known values, we get:
a = 13.28 N / 6 kg
a ≈ 2.213 m/s^2
Step 5: Calculate the speed of the block at the bottom of the ramp
Using the equations of motion, we can find the final velocity of the block when it reaches the bottom of the ramp.
The formula to calculate the final velocity (v) is:
v^2 = u^2 + 2 * a * s
Given:
- Initial velocity (u) = 0 m/s (as the block is released from rest)
- Distance traveled (s) = 3.5 m (length of the ramp)
- Acceleration (a) ≈ 2.213 m/s^2 (from Step 4)
Substituting the known values, we get:
v^2 = 0^2 + 2 * 2.213 m/s^2 * 3.5 m
v^2 ≈ 15.48 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 3.93 m/s
Therefore, the block's speed when it arrives at the bottom of the 3.5-meter long ramp is approximately 3.93 m/s.