First, let's analyze the block's motion when sliding down the ramp. By the conservation of mechanical energy, we have
1/2 * m * v^2 = m * g * h
where m is the mass of the block (2.25 kg), v is the velocity at the bottom of the ramp, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the ramp.
Now let's analyze the block's motion on the surface after leaving the ramp. The work done by the friction force is
W = F_friction * d = μk * m * g * d
where F_friction is the friction force, d is the width of the rough patch, and μk is the coefficient of kinetic friction. The work-energy principle states that the initial kinetic energy of the block plus the work done by the friction force equals the final kinetic energy:
1/2 * m * v_initial^2 + W = 1/2 * m * v_final^2
where v_initial is the initial velocity of the block after leaving the ramp and v_final is the final velocity after crossing the rough patch (4.00 m/s). We can plug in the expression for the work W from above to get
1/2 * m * v_initial^2 + μk * m * g * d = 1/2 * m * v_final^2.
Now we can substitute the expression for v_initial^2 from the conservation of mechanical energy equation and solve for h:
1/2 * m * (2 * g * h) + μk * m * g * d = 1/2 * m * v_final^2
m * g * h + μk * m * g * d = 1/2 * m * v_final^2
Dividing by m and rearranging gives
h = (1/2 * v_final^2 - μk * g * d) / g
Now we can plug in the given values to find h:
h = (1/2 * (4.00 m/s)^2 - 0.750 * 9.8 m/s² * 0.15 m) / (9.8 m/s²)
h ≈ 0.412 m
So the height of the ramp should be about 0.412 meters for the block's speed after crossing the rough patch to be 4.00 m/s.
As shown in the figure below, a 2.25-kg block is released from rest on a ramp of height h. When the block is released, it slides without friction to the bottom of the ramp, and then continues across a surface that is frictionless except for a rough patch of width 15.0 cm that has a coefficient of kinetic friction μk = 0.750. Find h such that the block's speed after crossing the rough patch is 4.00 m/s.
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