Starting from rest, a(n)9 kg block slides 9.7 m
down a frictionless ramp (inclined at 30◦
from
the floor) to the bottom. The block then
slides an additional 23.5 m along the floor
before coming to stop.
The acceleration of gravity is 9.8 m/s
2.Find the speed of the block at the bottom
of the ramp.
Answer in units of m/s
h = 9.7*sin30 = 4.85 m.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*4.85 = 95.06
V = 9.75 m/s.
To find the speed of the block at the bottom of the ramp, we can split the motion into two parts: the motion down the ramp and the motion along the floor.
First, let's find the speed of the block when it reaches the bottom of the ramp. We can use the equations of motion to solve this.
The initial velocity of the block is zero since it starts from rest. The acceleration of the block down the ramp is caused by gravity, and its value is g * sin(30), where g is the acceleration due to gravity (9.8 m/s^2) and sin(30) is the sine of the angle of the ramp.
Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement, we can solve for the final velocity at the bottom of the ramp.
v^2 = 0 + 2 * (g * sin(30)) * 9.7
v^2 = 2 * (9.8 * 0.5) * 9.7
v^2 = 94.46
v = sqrt(94.46)
v ≈ 9.72 m/s
Therefore, the speed of the block at the bottom of the ramp is approximately 9.72 m/s.